Solucao Calculo 5ed SEC.7.1a7.4
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(2) Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.1 Integration by Parts. 2. 8. Let u=x , dv=cos mxdx du=2xdx , v=. 1 sin mx . Then m. 1 2 2 x sin mx xsin mxdx ( * ). Next let U =x , dV =sin mxdx dU =dx , m m 1 1 1 1 1 V = cos mx , so xsin mxdx= xcos mx+ cos mxdx= xcos mx+ sin mx+C . 1 2 m m m m m Substituting for xsin mxdx in ( * ), we get 1 2 2 1 1 2 1 2 2 I= x sin mx xcos mx+ sin mx+C = x sin mx+ xcos mx sin mx+C , 1 2 2 3 m m m m m m m 2 where C= C . m 1 I= x cos mxdx= 2. 2 dx , v=x . Then 2x+1 2x (2x+1)1 ln (2x+1)dx =xln (2x+1) 2x+1 dx=xln (2x+1) 2x+1 dx 1 1 =xln (2x+1) 1 dx=xln (2x+1)x+ ln (2x+1)+C 2x+1 2 1 = (2x+1)ln (2x+1)x+C 2. 9. Let u=ln (2x+1) , dv=dx du=. 1. 10. Let u=sin x , dv=dx du=. 1. dx 2. t=1x , we get dt=2xdx , so 2. 1x xdx. 1. x. dx . Setting 2. 1x 2. = t. 1/2. . 1x 1. 1. , v=x . Then sin xdx=xsin x. 1 1 1/2 1/2 2 dt = 2t +C=t +C= 1x +C . 2 2. (. ). Hence, sin xdx=xsin x+ 1x +C . 2. 4. 11. Let u=arctan 4t , dv=dt du=. 2. dt=. 1+(4t). arctan 4t dt =t arctan 4t =t arctan 4t. 4t 1+16t. 2. 4 2. 1+16t 1 dt=t arctan 4t 8. dt , v=t . Then 32t 1+16t. 2. dt. 1 2 ln (1+16t )+C 8. 5. 12. Let u=ln p , dv= p dp 2.
(3) Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.1 Integration by Parts. du=. 1 1 6 1 6 1 5 1 6 1 6 5 dp , v= p . Then p ln pdp= p ln p p dp= p ln p p +C . p 6 6 6 6 36 1 dx , v=x . Then by Equation 2, x. 2. 13. First let u= ( ln x ) , dv=dx du=2ln x. 1 2 dx=x(ln x) 2 ln xdx . Next let U =ln x , dV =dx dU =1/xdx , V =x x to get ln xdx=xln x x ( 1/x ) dx=xln x dx=xln xx+C . Thus, I= (ln x) dx=x(ln x) 2 xln x 2. 2. 1. 2. (. I=x(ln x) 2 xln xx+C. 2. 1. ) =x(ln x) 2xln x+2x+C , where C=2C. 1. .. 14. Let u=t , dv=e dt du=3t dt , v=e . Then I= t e dt=t e 3t e dt . Integrate by parts twice 3. 2. t. 3 t. t. 3 t. 2 t. t. more with dv=e dt .. (. ). I = t 3et 3t 2et 6tet dt =t 3et 3t 2et+6tet 6et dt. (. ). = t 3et 3t 2et+6tet 6et+C= t 33t 2+6t6 et+C More generally, if p(t) is a polynomial of degree n in t , then repeated integration by parts shows that t / / / / / / n n t p ( t ) e dt= p(t) p (t)+ p (t) p (t)+ + ( 1 ) p ( ) (t) e +C . 2. 15. First let u=sin 3 , dv=e d du=3cos 3 d , v=. 1 2 e . Then 2. 1 2 3 2 2 e sin 3 e cos 3 d . Next let U =cos 3 , dV =e d 2 2 1 2 dU =3sin 3 d , V = e to get 2 1 2 3 2 2 e cos 3 d = 2 e cos 3 + 2 e sin 3 d . Substituting in the previous formula gives 1 2 3 2 9 2 1 2 3 2 9 I= e sin 3 e cos 3 e sin 3 d = e sin 3 e cos 3 I 2 4 4 2 4 4 13 1 2 3 2 1 2 4 I= e sin 3 e cos 3 +C . Hence, I = e ( 2sin 3 3cos 3 ) +C , where C= C . 1 4 2 4 13 13 1 I= e sin 3 d = 2. . . 1 sin 2 . Then 2 1 1 1 1 I= e cos 2 d = e sin 2 sin 2 e d = e sin 2 + e sin 2 d . 2 2 2 2 1 Next let U =e , dV =sin 2 d dU =e d , V = cos 2 , so 2 16. First let u=e. , dv=cos 2 d du=e d , v=. (. ). 3.
(4) Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.1 Integration by Parts. . 1 1 1 1 e cos 2 cos 2 e d = e cos 2 e cos 2 d . 2 2 2 2 1 1 1 1 1 1 1 So I= e sin 2 + e cos 2 I = e sin 2 e cos 2 I 2 2 2 2 2 4 4 5 1 1 I= e sin 2 e cos 2 +C 1 4 2 4 4 1 1 2 1 I= e sin 2 e cos 2 +C = e sin 2 e cos 2 +C . 1 5 2 4 5 5. e. (. sin 2 d =. ). 17. Let u=y , dv=sinh ydy du=dy , v=cosh y . Then ysinh ydy=ycosh y cosh ydy=ycosh ysinh y+C . sinh ay . Then a ysinh ay 1 ysinh ay ycosh aydy= a a sinh aydy= a cosh2 ay +C . a. 18. Let u=y , dv=cosh aydy du=dy , v=. 1 cos 3t . Then 3 1 1 1 1 0 t sin 3t dt= 3 t cos 3t 0 + 3 0 cos 3t dt= 3 0 + 9 sin 3t 0 = 3 .. 19. Let u=t , dv=sin 3t dt du=dt , v=. x. 2. x. 20. First let u=x +1 , dv=e dx du=2xdx , v=e x. 0 ( x +1 ) e 1. 2. x. , V =e 2. ). x 1. 1. 0. 0. x. 1. x 1. x. 0. x. 1. . By (6),. 1. x. x. + 2xe dx=2e +1+2 xe dx . Next let U =x , dV =e dx dU =dx. dx= x +1 e. . By (6) again, xe dx= xe. 0 ( x +1 ) e 1. (. 2. (. 1. ). 1. 0 1. 1 x. x 1. + e dx=e + e. 0. 0. 1. 1. 1. 1. 1. =e e +1=2e +1 . So. 0. 1. dx=2e +1+2 2e +1 =2e +14e +2=6e +3 . 2. 21. Let u=ln x , dv=x dx du=. 1 1 dx , v=x . By (6), x. 2. 1. ln x 2. x. ln x dx= x. 2 2 2. + x dx=. 1 1. 1 1 ln 2+ln 1+ 2 x. 22. Let u=ln t , dv= t dt du=dt/t , v=. 2. =. 1. 1 1 1 1 ln 2+0 +1= ln 2 . 2 2 2 2. 2 3/2 t . By Formula 6, 3. 4.
(5) Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.1 Integration by Parts. 1. 4. 4. 2 3/2 t ln t 3. t ln t dt=. 2 4 2 t dt= 8 ln 40 1 3 1 3. 2 2 3/2 t 3 3. 4. =. 1. 16 4 16 28 ln 4 ( 81 ) = ln 4 3 9 3 9. . dy. 23. Let u=y , dv=. 2y. 2y. =e. dy du=dy , v=. e. 0. y. 1. 2y. dy= . e. 1. 1 2y ye 2. +. 0. 1 2y e . Then 2. 1 1 2y 1 2 1 2y 1 1 2 1 2 1 1 3 2 e dy= e +0 e = e e + = e . 0 2 0 2 4 2 4 4 4 4. 2. 24. Let u=x , dv=csc xdx du=dx , v=cot x . Then /2 2 /2 /4xcsc xdx = xcot x /2/4+ /4cot xdx= 2 0+ 4 1+ ln sin x 1 1 1/2 +ln 1ln = +0ln 2 = + ln 2 = 4 2 4 2 4 1. dx. 25. Let u=cos x , dv=dx du=. /2. [see Exercise 5.5.]. /4. , v=x . Then 2. 1x 1/2. 1/2. 1. 1. I= cos xdx= xcos x 0. 1/2 0. . +. 2. 1x. 0. 1 Thus, I= + 6 2. 1. t. 1/2. dt=. 3/4. + 6. t. 1. =. 3/4. (. x. 3/4. xdx. 1 1 1/2 = + t 2 3 1 2. 2. dt , where t=1x dt=2xdx .. 3 1 +1 = ( +63 3 ) . 2 6 6. x. ). 26. Let u=x , dv=5 dx du=dx , v= 5 /ln 5 . Then 1. x5 dx = 0. x. x. x5 ln 5. 5 = ln 5 . 1 1. x. 5 5 1 dx= 0 0 ln 5 ln 5 ln 5 0. x. 5 ln 5. 1. =. 0. 5 ln 5. 5. ( ln 5). 2. +. 1. ( ln 5). 2. 4. ( ln 5). 2. cos x dx , v=sin x . Then sin x I= cos x ln ( sin x ) dx=sin x ln ( sin x ) cos xdx=sin x ln ( sin x ) sin x+C .. 27. Let u= ln ( sin x ) , dv=cos xdx du=. 5.
(6) Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.1 Integration by Parts. Another method: Substitute t=sin x , so dt=cos xdx . Then I= ln t dt=t ln tt+C (see Example 2) and so I=sin x ( ln sin x1 ) +C . 1. 28. Let u=arctan(1/x) , dv=dx du=. 2. 1. . 2. 1+(1/x) 3. arctan(1/x)dx =. xarctan. 1. =. 3. 1 x. 1. 3. + 1. dx=. dx. , v=x . Then. 2. x +1. x. x 2 1 2 x +1= 3 1 + ln (x +1) dx 6 4 2. 3 1. 3 1 3 1 4 3 1 + (ln 4ln 2)= + ln = + ln 2 6 6 6 4 2 2 2 2 2 2 w. w. 29. Let w=ln x dw=dx/x . Then x=e and dx=e dw , so 1 w w cos ( ln x ) dx = e cos wdw= 2 e ( sin w+cos w ) +C [by the method of Example 4] 1 = x sin ( ln x ) +cos ( ln x ) +C 2 r. 2. 30. Let u=r , dv=. 2. 4+r. dr du=2r dr , v= 4+r . By (6),. 2. 1. . r. 3. dr. 4+r. 0. 2. =. r. =. 5. 4+r. 2. 2. 1. 1. 2 2 r 4+r dr= 5 0 3 0 2. 2 3/2 2 (5) + (8)= 5 3 3. 1. 10 3. +. ( 4+r 2) 3/2. 1 0. 16 16 7 = 5 3 3 3. 5. ln x x 31. Let u=(ln x) , dv=x dx du=2 . By (6), dx , v= 5 x 2. 2. x (ln x) 4. 2. 4. 5. dx=. 1. x 2 ( ln x ) 5. 2. 2. 2. 4. 4. x x 32 2 2 ln xdx= ( ln 2 ) 02 ln xdx . 1 5 5 5 1 1. 4. 5. x 1 x Let U =ln x , dV = . dx dU = dx , V = 5 25 x 2. Then 1. 4. x ln xdx= 5. 5. x ln x 25. 2 2. 4. x 32 dx= ln 20 1 25 25 1. 5. x 125. 2. =. 1. 32 ln 2 25. 32 1 125 125. . 6.
(7) Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.1 Integration by Parts. 2. 32 2 (ln 2) 2 5. So x (ln x) dx= 4. 2. 1. 32 31 ln 2 25 125. 32 62 2 64 (ln 2) ln 2+ . 5 25 125. =. s. s. 32. Let u=sin (ts) , dv=e ds du=cos (ts)ds , v=e . Then t. t s. 0. 0. I= e sin (ts)ds= e sin (ts) + e cos (ts)ds=e sin 0e sin t+I . For I , let U =cos (ts) , dV =e ds t s. s. 0. 0. t. t. t s. 0. 0. s. 1 t. 1 0. dU =sin (ts)ds , V =e . So I = e cos (ts) e sin (ts)ds=e cos 0e cos tI . Thus, s. s. 1. t. t. I=sin t+e cos tI 2I=e cos tsin t I=. 1 t e cos tsin t . 2. (. ). 33. Let w= x , so that x=w and dx=2wdw . Thus, sin x dx= 2wsin wdw . Now use parts with u=2w , dv=sin wdw , du=2dw , v=cos w to get 2. 2wsin wdw =. 2wcos w+ 2cos wdw=2wcos w+2sin w+C. = 2 x cos x +2sin x +C=2 ( sin x x cos x ) +C 34. Let w= x , so that x=w and dx=2wdw . Thus, e 4. 2. dv=e dw , du=2dw , v=e to get e 2wdw= 2we w. 2 w. w. 1 w 2. x. dx= e 2wdw . Now use parts with u=2w , 2 w 1. (. ). 2 e dw=4e 2e2 e e =2e .. 1. 1. 2 w. 2. 1. 2. 2. 2. 35. Let x= , so that dx=2 d . Thus, . . 3. ( 2) d =. cos . /2. . ( 2) 12 (2 d )= 12 /2xcos x}{dx} . Now use parts with u=x ,. 2. /2. cos . dv=cos xdx , du=dx , v=sin x to get . 1 1 xcos xdx = 2 /2 2 =. xsin x. . . sin xdx. /2. /2. 1 1 ( sin +cos ) 2 2. =. 1 xsin x+cos x 2. sin +cos 2 2 2. =. /2. 1 1 ( 01) 2 2. 1 1+0 = 2 2 4. 36.. x e. 2. 5 x. 2. 2 2 x. ( )e. dx = x. xdx= t e. 2 t. 1 1 2 dt [ where t=x dt=xdx ] 2 2 2. 1 4 2 1 2 t x t 2t+2 e +C [ by Example 3] = x 2x +2 e +C = 2 2. (. ). (. ). 7.
(8) Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.1 Integration by Parts. 37. Let u=x , dv=cos xdx du=dx , v= ( sin x ) / . Then sin x sin x xsin x xcos xdx=x dx= + cos 2 x +C . We see from the graph that this is reasonable, since F has extreme values where f is 0 .. 1 2 5/2 dx , v= x . Then x 5 2 3/2 2 5/2 2 2 5/2 2 5/2 3/2 = x ln x x dx= x ln x x +C x ln xdx 5 5 5 5 4 5/2 2 5/2 = x ln x x +C. 25 5 We see from the graph that this is reasonable, since F has a minimum where f changes from negative to positive. 38. Let u=ln x , dv=x. 3/2. dx du=. x. x. 39. Let u=2x+3 , dv=e dx du=2dx , v=e . Then. (2x+3)e. dx=(2x+3)e 2 e dx=(2x+3)e 2e +C= ( 2x+1 ) e +C . We see from the graph that this is reasonable, since F has a minimum where f changes from negative to positive. x. x. x. x. x. x. 8.
(9) Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.1 Integration by Parts. 2. 2. 40. x e dx= x xe dx=I . 3 x. 2. x. 2. 2. 1 x Let u=x , dv=xe dx du=2xdx , v= e . Then 2 2 2 2 2 2 1 2 x 1 2 x 1 x 1 x 2 x I= x e xe dx= x e e +C= e x 1 +C . We see from the graph that this is 2 2 2 2 reasonable, since F has a minimum where f changes from negative to positive. 2. x. (. ). 1 1 x sin 2x cos xsin x+ 1dx= +C . 2 2 2 4 1 3 1 3 3 4 3 2 3 sin 2x+C . (b) sin xdx= cos xsin x+ sin xdx= cos xsin x+ x 4 4 4 8 16 41. (a) Take n=2 in Example 6 to get sin xdx= 2. 42. (a) Let u=cos. cos. n. n1. x , dv=cos xdx du=(n1)cos. n2. xsin xdx , v=sin x in (2):. n1 n2 2 xdx = cos xsin x+(n1) cos xsin xdx. (. ). = cos n1 xsin x+(n1) cos n2 x 1cos 2 x dx = cos n1 xsin x+(n1) cos n2 xdx(n1) cos nxdx Rearranging terms gives n cos xdx=cos n. cos. n. xdx=. n1. xsin x+(n1) cos. n2. xdx or. 1 n1 n1 n2 cos xsin x+ cos xdx n n. 1 1 x sin 2x cos xsin x+ 1dx= + +C . 2 2 2 4 1 3 1 3 3 4 3 2 3 sin 2x+C (c) cos xdx= cos xsin x+ cos xdx= cos xsin x+ x+ 4 4 4 8 16 (b) Take n=2 in part (a) to get cos xdx= 2. 43. (a) From Example 6, sin xdx= n. 1 n1 n1 n2 cos xsin x+ sin xdx . Using (6), n n. 9.
(10) Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.1 Integration by Parts. /2. 0. /2. n1. n. cos xsin x n1 /2 n2 sin xdx = + sin xdx 0 n n 0 n1 /2 n2 n1 /2 n2 = (00)+ sin xdx= sin xdx n 0 n 0. /2 2 2 /2 2 = . (b) Using n=3 in part (a), we have sin xdx= sin xdx= cos x 0 0 3 0 3 3 /2 4 /2 3 4 2 8 5 Using n=5 in part (a), we have sin xdx= sin xdx= = . 0 5 0 5 3 15 (c) The formula holds for n=1 (that is, 2n+1=3 ) by (b). Assume it holds for some k 1 . Then /2 2 4 6 ( 2k ) 2k+1 0 sin xdx= 3 5 7 ( 2k+1 ) . By Example 6, /2 2k+3 2k+2 2 4 6 ( 2k ) 2k+2 /2 0 sin xdx = 2k+3 0 sin 2k+1xdx= 2k+3 3 5 7 2k+1 ( ) 2 4 6 (2k)[2 ( k+1 ) ] = , 3 5 7 (2k+1)[2 ( k+1 ) +1] so the formula holds for n=k+1 . By induction, the formula holds for all n 1 . /2. 3. 44. Using Exercise 43 (a), we see that the formula holds for n=1 , because /2 1 /2 1 2 /2 1 0 sin xdx= 2 0 1dx= 2 x 0 = 2 2 . /2 1 3 5 (2k1) 2k Now assume it holds for some k 1 . Then sin xdx= . By Exercise 43 0 2 4 6 ( 2k ) 2 (a), /2 2 k+1 2k+1 1 3 5 (2k1) 2k+1 /2 0 sin ( ) xdx = 2k+2 0 sin 2k xdx= 2k+2 2 4 6 2k 2 ( ) 1 3 5 (2k1)(2k+1) = , 2 4 6 (2k)(2k+2) 2 so the formula holds for n=k+1 . By induction, the formula holds for all n 1 . n. n1. 45. Let u=(ln x) , dv=dx du=n(ln x). (ln x). n. dx=x(ln x) nx(ln x) n. n. n1. x. 46. Let u=x , dv=e dx du=nx. (. 2. 2 n. 47. Let u= x +a. ). ( dx/x ) , v=x . By Equation 2,. ( dx/x ) =x(ln x) n (ln x) dx .. n1. n. n1. dx , v=e . By Equation 2, x e dx=x e n x x. (. 2. n x. n x. n1 x. e dx .. 2 n1. , dv=dx du=n x +a. ). 2xdx , v=x . Then 10.
(11) Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.1 Integration by Parts. 2 n. ) 2n x2 ( x2+a2) n1 dx 2 2 n 2 2 n 2 2 2 n1 2 2 2 2 = x ( x +a ) 2n ( x +a ) dxa ( x +a ) dx since x = ( x +a ) a 2 2 n 2 2 n 2 2 2 n1 ( 2n+1 ) ( x +a ) dx=x ( x +a ) +2na ( x +a ) dx , and 2 2 n 2 x ( x +a ) 2na 2 2 n 2 2 n1 ( x +a ) dx= 2n+1 + 2n+1 ( x +a ) dx [ provided 2n+1 0 ]. ( x +a 2. 48. Let u=sec. sec. n. ). (. 2 n. 2. dx = x x +a. n2. 2. x , dv=sec xdx du= ( n2 ) sec. n3. xsec xtan xdx , v=tan x . Then by Equation 2,. n2 n2 2 xdx = tan xsec x ( n2 ) sec xtan xdx. (. ). = tan xsec n2 x ( n2 ) sec n2 x sec 2 x1 dx = tan xsec n2 x ( n2 ) sec nxdx+ ( n2 ) sec n2 xdx so ( n1 ) sec xdx=tan xsec n. sec. n. xdx=. tan xsec n1. n2. x. +. n2. x+ ( n2 ) sec. n2. xdx . If n1 0 , then. n2 n2 sec xdx . n1. 49. Take n=3 in Exercise 45 to get (ln x) dx=x ( ln x ) 3 (ln x) dx=x(ln x) 3x(ln x) +6xln x6x+C [ by Exercise 13 ]. Or : Instead of using Exercise 13 , apply Exercise 45 again with n=2 . 3. 3. 2. 3. 2. 50. Take n=4 in Exercise 46 to get. x e. 4 x. (. ). 4 x 3 x 4 x 3 2 x dx = x e 4 x e dx=x e 4 x 3x +6x6 e +C. (. [by Exercise 14 ]. ). = ex x44x3+12x224x+24 +C Or: Instead of using Exercise 14 , apply Exercise 46 with n=3 , then n=2 , then n=1 . 0.4x. 51. Area = xe 5. 0.4x. dx . Let u=x , dv=e. 0. 0.4x. du=dx , v=2.5e. dx. . Then. area = 2.5xe0.4x 5+2.5 5e0.4xdx 0. 0. 2. 0.4x 5. =12.5e +0+2.5 2.5e 2. 2. 0 2. =12.5e 6.25(e 1)=6.2518.75e. or. 25 75 2 e 4 4 11.
(12) Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.1 Integration by Parts. 52. The curves y=xln x and y=5ln x intersect when xln x=5ln x xln x5ln x=0 ( x5) ln x=0 ; that is, when x=1 or x=5 . For 1<x<5 , we have 5ln x>xln x since ln x>0 . Thus, area 5 5 1 2 = ( 5ln xxln x ) dx= (5x)ln x dx . Let u=ln x , dv= ( 5x ) dx du=dx/x , v=5x x . Then 1 1 2 5 5 5 1 2 1 25 1 1 2 area = (ln x) 5x x 5x x dx=(ln 5) 0 5 x dx 1 1 1 x 2 2 2 2 =. 1 2 25 ln 5 5x x 4 2. 5. =. 1. 25 ln 5 2. 25. 25 4. 5. 1 4. =. 25 ln 514 2. 2. 53. The curves y=xsin x and y=(x2) intersect at a 1.04748 and b 2.87307 , so area =. b a. 2. xsin x(x2) dx. 1 3 b [ by Example 1] (x2) 3 a. 2.813580.63075=2.18283. = xcos x+sin x. 54. The curves y=arctan3x and y=x/2 intersect at x= a 2.91379 , so area = a arctan3x 1 x dx=2 a arctan3x 1 x dx a 0 2 2 1 2 1 2 =2 xarctan3x ln (1+9x ) x 4 6. 2(1.39768)=2.79536 .. a 0. [ see Example 5]. 12.
(13) Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.1 Integration by Parts. 2 sin ( x/2) . 2 2 0 4 cos . 55. V = 2 xcos ( x/2)dx . Let u=x , dv=cos ( x/2)dx du=dx , v= 1. 0. x 2 8 8 . =4+ (01)=4 . V =2. 1. 2 2 0 . 2 xsin . 0sin 1. x 2. dx=2. x 2. 1 0. 56.. (. x. Volume = 2 x e e 1. 0. x. ) dx=2 10 ( xexxex) dx x. =2. 0xe. =2. ( xexex) ( xexex). 1. x. dx xe dx [ both integrals by parts] 1. 0. 1. =2 [2/e0]=4 /e. 0. x. x. x. 57. Volume = 2 (1x)e dx . Let u=1x , dv=e dx du=dx , v=e 0. 1. x. V =2 (1x)(e ) x 0. =2 xe. 0. x. x. x 0. 2 e dx=2 (x1)(e )+e. 1. 0. 1. .. 1. =2 (0+e)=2 e. 1. 58. Volume. . = 2 y ln ydy=2 1. =2. 1 2 y (2ln y1) 4. 1 2 1 2 y ln y y 2 4. 1. 2. 1. . =2. (2ln 1) (01) 4 4. 2. 59. The average value of f (x)=x ln x on the interval 1,3 is f 2. Let u=ln x , dv=x dx du=(1/x)dx , v=. 3. = ln + 2 2 3. =. ave. 1 3 2 1 x ln xdx= I . 31 1 2. 1 3 x . So 3 13.
(14) Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.1 Integration by Parts. 3 31 2 1 3 1 3 3 1 I= x ln x x dx= ( 9ln 30 ) x =9ln 3 3 1 1 3 1 3 9 9 1 1 26 9 13 9ln 3 = ln 3 Thus, f = I= . ave 2 2 9 2 9. =9ln 3. 26 . 9. 60. The rocket will have height H= v(t)dt after 60 seconds. 60 0. H =. 0. 60. mrt m. gtv ln e. dt=g. 60. 1 2 t 2. v. 0. 0 ln (mrt)dt 0 ln mdt 60. e. 60. = g(1800)+v (ln m)(60)v ln (mrt)dt 60. e 0. e. 1 (r)dt , v=t . Then mrt 60 60 60 rt dt=60ln (m60r)+ tln (mrt) + 0 0 mrt 0. Let u=ln (mrt) , dv=dt du=. 0 ln (mrt)dt = 60. = 60ln (m60r)+ t. m ln (mrt) r. 1+. m mrt. dt. 60 0. m m ln (m60r)+ ln m r r m m So H=1800g+60v ln m60v ln (m60r)+60v + v ln (m60r) v ln m . Substituting g=9.8 , e e e r e r e m=30 , 000 , r=160 , and v =3000 gives us H 14 , 844 m. = 60ln (m60r)60. e. 2 w. 61. Since v(t)>0 for all t , the desired distance is s(t)= v(w)dw= w e dw . t. t. 0. w. 2. w. First let u=w , dv=e dw du=2wdw , v=e w. w. Next let U =w , dV =e dw dU =dw , V =e 2 t. w t. s(t) = t e +2. we. t w. 0 2 w t. . Then s(t)= w e. t. 0. 0. . Then. 2 t. t. w t. + e dw =t e +2 te +0+ e. 0. w. +2 we dw .. 0. 0. = t 2et+2 tet et+1 =t 2et 2tet 2et+2. (. ). = 2et t 2+2t+2 meters. (. ). 62. Suppose f (0)=g(0)=0 and let u= f (x) , dv=g. 0 f (x)g a. / /. a. a. 0. 0. / /. /. /. (x)dx du= f (x)dx , v=g (x) . Then. (x)dx= f (x)g (x) f (x)g (x)dx= f (a)g (a) f (x)g (x)dx . Now let U = f (x) , /. /. /. /. a. /. /. /. 0. 14.
(15) Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.1 Integration by Parts. /. / /. dV =g (x)dx dU = f. 0 f a. /. (x)dx and V =g(x) , so a. a. 0. 0. (x)g (x)dx= f (x)g(x) f /. /. / /. (x)g(x)dx= f (a)g(a) f. a. / /. 0. 4. / /. 1 4. 4. 1. 1. / /. 0. Combining the two results, we get f (x)g 63. For I= xf. a. /. / /. (x)dx , let u=x , dv= f. (x)g(x)dx .. (x)dx= f (a)g (a) f (a)g(a)+ f /. a. /. / /. 0. (x)g(x)dx .. /. (x)dx du=dx , v= f (x) . Then. I= xf (x) f (x)dx=4 f (4)1 f (1)[ f (4) f (1)]=4 31 5(72)=1255=2 . /. /. /. We used the fact that f. / /. /. is continuous to guarantee that I exists. /. 64. (a) Take g(x)=x and g (x)=1 in Equation 1. (b) By part (a), f (x)dx=bf (b)a f (a) x f (x)dx . Now let y= f (x) , so that x=g(y) and b. b. a. a. dy= f (x)dx . Then x f (x)dx= b. /. /. /. a. f ( b) f ( a). g(y)dy . The result follows.. (c) Part (b) says that the area of region ABFC is af (a). = bf (b). . f (( a )) g(y)dy f b. = ( area of rectangle OBFE ) ( area of rectangle OACD ) ( area of region DCFE ). 1. x. (d) We have f (x)=ln x , so f (x)=e , and since g= f. 1ln xdx=eln e1ln 1 ln 1e. ln e y. e. dy=e e dy=e e 1 y 0. 1. y. , we have g(y)=e . By part (b),. y 1. =e(e1)=1 .. 0. 65. Using the formula for volumes of rotation and the figure, we see that Volume = b dy a dy g(y) dy= b d a c g(y) dy . Let y= f (x) , which gives dy= f (x)dx d. 2. 0. c. 0. 2. d. 2. 2. 2. c. d. 2. /. c. and g(y)=x , so that V = b d a c x f (x)dx . Now integrate by parts with u=x , and 2. 2. b 2. /. 2. a. /. dv= f (x)dx du=2xdx , v= f (x) , and 15.
(16) Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.1 Integration by Parts. ax. b 2. b. f (x)dx= x f (x) 2x f (x)dx=b f (b)a f (a) 2x f (x)dx , but f (a)=c and f (b)=d /. 2. b. a. 2. b. 2. a. a. V = b d a c b da c 2xf (x)dx = 2 xf (x)dx . 2. 2. 2. 2. b. b. a. a. 2n+2 2n+1 2n x
(17) sin x
(18) sin x . So by the second , 0
(19) sin x
(20) 1 , so sin 2
(21) I
(22) I . Comparison Property of the Integral, I 66. (a) We note that for 0
(23) x
(24). 2n+2. 2n+1. 2n. (b) Substituting directly into the result from Exercise 44 , we get I. 1 3 5 2(n+1)1 2 2 4 6 2(n+1) 2(n+1)1 2n+1 2n+2 = = = 1 3 5 (2n1) 2(n+1) 2n+2 I 2n 2 4 6 ( 2n ) 2. (c) We divide the result from part (a) by I I. 2n+2. I. 2n.
(25). I. 2n+1. I. 2n.
(26). I I. 2n. 2n. . The inequalities are preserved since I. . Now from part (b), the left term is equal to. 2n. 2n. is positive:. 2n+1 , so the expression becomes 2n+2. I 2n+1 I 2n+1 2n+1 2n+1
(27)
(28) 1 . Now lim =lim 1=1 , so by the Squeeze Theorem, lim =1 . 2n+2 I n 2n+2 n n I 2n. 2n. (d) We substitute the results from Exercises 43 and 44 into the result from part (c): 2 4 6 ( 2n ) I 3 5 7 ( 2n+1 ) 2n+1 1 = lim =lim n I n 1 3 5 ( 2n1 ) 2n 2 4 6 ( 2n ) 2 2 4 6 ( 2n ) 2 4 6 ( 2n ) 2 = lim 3 5 7 ( 2n+1 ) 1 3 5 ( 2n1 ) n 2 2 4 4 6 6 2n 2n 2 = lim 1 3 3 5 5 7 2n1 2n+1 n Multiplying both sides by. [rearrange terms]. gives us the Wallis product : 2. 2 2 4 4 6 6 = 2 1 3 3 5 5 7 (e) The area of the k th rectangle is k . At the 2n th step, the area is increased from 2n1 to 2n by 16.
(29) Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.1 Integration by Parts. 2n , and at the ( 2n+1 ) th step, the area is increased from 2n to 2n+1 by 2n1 2n+1 2n multiplying the height by . These two steps multiply the ratio of width to height by and 2n 2n1 1 2 2 4 4 6 6 2n respectively. So, by part (d), the limiting ratio is = . = (2n+1)/(2n) 2n+1 1 3 3 5 5 7 2 multiplying the width by. 17.
(30) Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.2 Trigonometric Integrals. 1.. sin. 3. ( ). ). (. 2. ). (. 2. xcos xdx = sin xcos xsin xdx= 1cos x cos xsin xdx= 1u u (du) 1 5 1 3 1 1 2 2 4 2 5 3 = u 1 u du= u u du= u u +C= cos x cos x+C 5 3 5 3 2. 2. 2. (. ). (. 2. 2. 2. 2.. sin. 6. (. ). ). xcos xdx = sin xcos xcos xdx= sin x 1sin x cos xdx= u 1u du 1 7 1 9 1 1 6 8 7 9 = u u du= u u +C= sin x sin x+C 7 9 7 9 3. 6. 2. (. 6. 2. 6. ). 3. 3 /4. . 5. 3 /4. 3. sin xcos xdx =. /2. . 5. 3 /4. 2. sin xcos xcos xdx=. /2 s =. 5. . u. 5. (. 2. 2 /2. 1. 1. 1/8 1/16 6 8. 1 1 6 8. . =. 2 /2. 1 6 1 8 u u 6 8. ( 1u ) du= ( u5u7) du= 2. ). sin x 1sin x cos xdx. /2. 2 /2. =. . 1. 11 384. 4. /2. cos. 5. /2. (. xdx = cos x. 0. 2. ). 2. /2. 0 1. (. ). 1. 2. 0. (. = 12u +u 2. 4. (. cos xdx= 1sin x cos xdx= 1u 2. ) du=. 0. u. 2 2. ). du. 0 1. 2 3 1 5 u+ u 3 5. = 1. 0. 2 1 + 3 5. 0=. 8 15. 5.. cos. 5. 4. (. )2 ). (. 2 2 4. ). xsin xdx = cos xsin xcos xdx= 1sin x sin xcos xdx= 1u u du 1 5 2 7 1 9 2 4 4 4 6 8 = 12u +u u du= u 2u +u du= u u + u +C 5 7 9 1 2 1 5 7 9 = sin x sin x+ sin x+C 5 7 9 4. (. 4. ). (. 2. 4. 6. 1.
(31) Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.2 Trigonometric Integrals. sin. 1 2 2 (mx)dx = 1cos mx sin mxdx= m 1u du [ u=cos mx , du=msin mxdx ] 1 1 3 1 1 3 = u u +C= cos mx cos mx +C m 3 m 3 1 1 3 = cos mx cos mx+C 3m m. (. 3. ). (. ). 7. /2. 0. /2. cos d = 2. /2 0. 0. 1 2. = 8. . 1 (1+cos 2 )d [ halfangle identity] 2 + /2. sin (2 )d = 2. 0. /2. 1 sin 2 2. =. 0. +0 ( 0+0 ) 2. 1 2. 1 1 (1cos 4 )d = 2 2. . 1 sin 4 4. =. /2 0. =. 1 2. 4 0 ( 00 ) 2. =. 4. 9. . 0 sin. . 4. (3t)dt = =. 0. 2. 1 2 (12cos 6t+cos 6t)dt 0 4 0 1 1 3 1 12cos 6t+ (1+cos 12t) dt= 2cos 6t+ cos 12t dt 2 4 0 2 2 1 (1cos 6t) 2. sin (3t) dt=. 1 4 0. 1 = 4. . 2. 2. 3 1 1 t sin 6t+ sin 12t 2 3 24. 0. =. dt=. 3 0+0 ( 00+0 ) 2. 1 4. =. 3 8. 10. . 0 cos. . 6. . d = (cos ) d = 2. 3. 0. =. 1 8. +. 3 sin 2 2. 1 (1+cos 2 ) 2. 0 0. +. 1 8 0. 3 1 1 + sin 4 = + 16 4 8 3 5 = + +0= 8 16 16. d =. 1 2 3 (1+3cos 2 +3cos 2 +cos 2 )d 8 0. 3 1 2 (1+cos 4 ) d + (1sin 2 )cos 2 d 2 8 0. 0. 3. +. 1 0 2 (1u ) 8 0. 1 du 2. [ u=sin 2 , du=2cos 2 d ]. 11.. 2.
(32) Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.2 Trigonometric Integrals. (1+cos ). 2. 1 2 d = (1+2cos +cos )d = +2sin + 2 (1+cos 2 )d 1 1 3 1 = +2sin + + sin 2 +C= +2sin + sin 2 +C 2 4 2 4 1 1 1 (1+cos 2x)dx= x+ sin 2x , so 2 2 4 1 1 1 2 1 1 2 1 x+ sin 2x dx= x + xsin 2x x + cos 2x+C 2 4 2 4 4 8. 12. Let u=x , dv=cos xdx du=dx , v= cos xdx= 2. xcos. 2. 2. 1 1 x+ sin 2x 4 2 1 2 1 1 = x + xsin 2x+ cos 2x+C 4 4 8. xdx =x. 13. /4. 0. 4. 2. 2 1 sin 2x dx 0 0 2 1 /4 1 /4 2 1 /4 2 2 = (1cos 2x)sin 2xdx= sin 2xdx sin 2xcos 2xdx 8 0 8 0 8 0 /4. sin xcos xdx =. /4. sin x(sin xcos x) dx= 2. 2. 1 (1cos 2x) 2. 1 1 1 /4 3 (1cos 4x)dx sin 2x 3 16 16 0 1 1 1 = 0 = (3 4) 4 16 3 192 =. /4 0. =. 1 16. 1 1 3 sin 4x sin 2x 4 3. x. /4 0. 14. /2. 0. /2. sin xcos xdx = 2. 2. 0. /2 1 1 1 /2 2 2 2 2 4sin xcos x dx= (2sin xcos x) dx= sin 2xdx 0 4 4 4 0. (. ). 1 /2 1 1 /2 1 (1cos 4x)dx= (1cos 4x)dx= 8 0 8 4 0 2 1 = = 2 8 16. x. =. 1 sin 4x 4. /2 0. 15.. sin. 3. (. ). (. ). (. x cos x dx = 1cos x cos x sin xdx= 1u u ( du ) = u 2 7/2 2 3/2 2 7/2 2 3/2 = u u +C= ( cos x ) ( cos x ) +C 7 3 7 3 2 2 3 = cos x cos x cos x +C 7 3 2. 2. 1/2. 5/2. u. 1/2. ) du. 3.
(33) Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.2 Trigonometric Integrals. 16. Let u=sin . Then du=cos d and. cos cos. (sin )d = cos u du= (cos u) cos udu= (1sin u) cos u du 2 4 = (12sin u+sin u)cos u du=I 5. 5. 2. 2. 2. 2. Now let x=sin u . Then dx=cos udu and 2 3 1 5 2 1 2 4 3 5 I = (12x +x )dx=x 3 x + 5 x +C=sin u 3 sin u+ 5 sin u+C 2 1 3 5 =sin (sin ) sin (sin )+ sin (sin )+C 3 5 17.. (. 3. 2. ). cos xtan xdx = sin x dx= 1u ( du ) = 1 +u du u cos x u 1 2 1 2 =ln u + u +C= cos xln cos x +C 2 2 2. 3. 18.. cot. sin d =. 5. 4. 5. cos 5. sin . 5. (. 4. 2. cos cos 1sin sin d = d = cos d = sin sin sin 4. ) 2 cos d. 2 2 2 4 ( 1u ) 12u +u du= du= =. 1 3 2u+u du u u u 1 2 1 4 2 4 =ln u u + u +C=ln sin sin + sin +C 4 4. 19. 1sin x cos x dx = ( sec xtan x ) dx=ln sec x+tan x ln sec x +C =ln ( sec x+tan x ) cos x +C=ln 1+sin x +C =ln ( 1+sin x ) +C since 1+sin x 0. by (1) and the boxed formula above it. Or:. . (. ). 2 1sin x 1sin x 1sin x dx cos xdx 1+sin x dx = dx= = cos x cos x 1+sin x cos x ( 1+sin x ) 1+sin x 4.
(34) Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.2 Trigonometric Integrals. dw [ where w=1+sin x , dw=cos xdx ] w =ln w +C=ln 1+sin x +C=ln ( 1+sin x ) +C. =. 20. cos xsin 2xdx=2 cos xsin x dx=2 u du= 2. 3. 3. 1 4 1 4 u +C= cos x+C 2 2. 1 2 1 2 u +C= tan x+C . 2 2 1 2 1 2 2 Or: Let v=sec x , dv=sec xtan xdx . Then sec xtan xdx= vdv= v +C= sec x+C . 2 2 21. Let u=tan x , du=sec xdx . Then sec xtan xdx= udu= 2. 2. 22. /2. 0. /4. sec (t/2)dt = 4. 4. sec x(2dx) [ x=t/2 , dx=. 0. /4 1 2 2 dt ] =2 sec x(1+tan x)dx 0 2. =2 (1+u )du [ u=tan x , du=sec xdx ] =2 1. 2. 2. 0. (. u+. 1 3 u 3. 1. =2. 0. 1+. 1 3. =. 8 3. ). 23. tan xdx= sec x1 dx=tan xx+C 2. 2. 1 3 tan xtan x+x+C 3 (Set u=tan x in the first integral and use Exercise 23 for the second.). (. ). 24. tan xdx= tan x sec x1 dx= tan xsec xdx tan xdx= 4. 2. 2. 2. 2. 2. 25.. sec. t dt = sec t sec t dt= (tan t+1) sec t dt= (u +1) du 1 5 2 3 1 4 2 5 2 3 = (u +2u +1)du= u + u +u+C= tan t+ tan t+tan t+C 5 3 5 3. 6. 4. 2. 2. 2. 2. 2. 2. 26. /4. 0. /4. sec tan d = 4. 4. (tan +1)tan sec d = (u +1)u du. 0. 2. 4. 1. 2. 2. 4. 0. = (u +u )du= 1. 6. 4. 0. 1 7 1 5 u+ u 7 5. 1. =. 0. 1 1 12 + = 7 5 35. 27. /3. 0. /3. tan xsec xdx = 5. 4. 0. 5. 2. 2. tan x(tan x+1)sec xdx 5.
(35) Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.2 Trigonometric Integrals. = =. 3 5 2. u (u +1)du. 0 3. 7. 0. 3. 1 8 1 6 u+ u 8 6. 5. (u +u )du=. =. 0. 81 27 81 9 81 36 117 + = + = + = 8 6 8 2 8 8 8. Alternate solution: /3. 0. /3. tan xsec xdx = 5. 4. /3. tan xsec xsec xtan xdx=. 0. 4. 3. 2. 2. 3. (sec x1) sec xsec xtan xdx. 0. = (u 1) u du 2. 2. 2 3. 1. = (u 2u +1)u du= (u 2u +u )du 2. 4. 2. 2. 3. 1. 5. 3. 1. 1 8 1 6 1 4 u u+ u 3 4 8. =. 7. 2. = 32. 1. 64 +4 3. 1 1 1 + 8 3 4. =. 117 8. 28.. tan. (2x)sec (2x)dx = tan (2x)sec (2x) sec (2x)tan (2x)dx 2 4 1 = (u 1)u ( du) [ u=sec (2x),du=2sec (2x)tan (2x)dx ] 2 1 1 7 1 5 1 1 6 4 7 5 = (u u )du= u u +C= sec (2x) sec (2x)+C 2 14 10 14 10. 3. 5. 2. 4. 29.. tan. 3. (. ). xsec xdx = tan xsec xtan xdx= sec x1 sec xtan xdx 2. 2. = (u 1)du 1 3 1 3 = u u+C= sec xsec x+C 3 3 2. 30. /3. 0. 5. 6. /3. tan xsec xdx = = =. 0. 5. 3 5. (. u 1+u. 0 3 0. /3. tan xsec xsec xdx= 4. 2 2. ). 2. 0. 5. (. 2. )2. du [ u=tan x , du=sec xdx ] =. ( u5+2u7+u9) du=. 2. tan x 1+tan x sec xdx 3 5. 2. 1 6 1 8 1 10 u+ u+ u 6 4 10. 0 3. 0. (. 2. 4. ). u 1+2u +u du. =. 27 81 243 981 + + = 6 4 10 20 6.
(36) Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.2 Trigonometric Integrals. Alternate solution: /3. 0. 5. /3. 6. tan xsec xdx =. /3. tan xsec xsec xtan xdx=. 0. 4. 5. 0. ( sec 2x1) 2sec 5xsec xtan xdx. (1 2 ) 2 5 2 4 2 9 2 5 7 5 = ( u 2u +1 ) u du= ( u 2u +u ) du 1 1. = u 1 u du [ u=sec x , du=sec xtan xdx ] 2. 2. 1 10 1 8 1 6 u u+ u 10 4 6. =. =. 1. 512 32 64+ 5 3. . 1 1 1 + 10 4 6. =. 981 20. 31.. tan. 5. (. )2. xdx = sec x1 tan xdx= sec xtan xdx2 sec xtan xdx+ tan xdx 2. 4. 2. = sec xsec xtan xdx2 tan xsec xdx+ tan xdx 1 1 4 2 4 2 = sec xtan x+ln sec x +C [ or sec xsec x+ln sec x +C ] 4 4 3. 2. 32.. tan. 6. (. ). aydy = tan ay sec ay1 dy= tan aysec aydy tan aydy 1 5 2 2 = tan ay tan ay sec ay1 dy 5a 1 5 2 2 2 = tan ay tan aysec aydy+ sec ay1 dy 5a 1 1 1 5 3 = tan ay tan ay+ tan ayy+C 5a 3a a 4. 2. 4. (. 2. 4. ). (. ). 33.. . 3. tan 4. cos . d = tan 3 sec 4 d = tan 3 (tan 2 +1) sec 2 d = u (u +1)duu=tan , du=sec d ] 1 6 1 4 1 1 5 3 6 4 = (u +u )du= u + u +C= tan + tan +C 6 4 6 4 3 2. 2. 34.. tan. 2. (. ). xsec xdx = sec x1 sec xdx= sec xdx sec xdx 2. 3. 7.
(37) Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.2 Trigonometric Integrals. 1 ( sec xtan x+ln sec x+tan x ) ln sec x+tan x +C [ by Example 8 and (1)] 2 1 = ( sec xtan xln sec x+tan x ) +C 2 =. /2. 35. . /2. 2 /2 ( csc x1 ) dx= cot xx = /6 /6. cot xdx= 2. /6. 0. 2. 3. 6. = 3. 3. 36. /2. /4cot. 3. /2. xdx =. (. ). /2. cot x csc x1 dx=. /4. = . 2. 1 2 cot xln sin x 2. /2. cot xcsc xdx 2. /4. /4. /2. = ( 0ln 1 ) . /4. cos x dx sin x. 1 1 ln 2 2. =. 1 1 1 +ln = (1ln 2) 2 2 2. 37.. cot. csc d = cot csc csc cot d = (csc 1)csc csc cot d. 3. 2. 3. 2. 2. 2. = (u 1)u (du) [ u=csc , du=csc cot d ] 1 3 1 5 1 1 2 4 3 5 = (u u )du= u u +C= csc csc +C 3 5 3 5 2. 2. 38.. csc. 4. xcot xdx = cot x(cot x+1)csc xdx 6. 6. 2. 2. = u (u +1) (du) [ u=cot x , du=csc xdx ] 1 9 1 7 1 1 8 6 9 7 = (u u )du= u u +C= cot x cot x+C 9 7 9 7 6. 2. 2. 2. csc x ( csc xcot x ) csc xcot x+csc x 39. I= csc xdx= dx= dx . Let u=csc xcot x csc xcot x csc xcot x. (. ). du= csc xcot x+csc x dx . Then I= du/u=ln u =ln csc xcot x +C . 2. 40.. csc. 4. xcot xdx = cot x(cot x+1)csc xdx 6. 6. 2. 2. = u (u +1) (du) [ u=cot x , du=csc xdx ] 6. 2. 2. 8.
(38) Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.2 Trigonometric Integrals. = (u u )du= 8. 6. 1 9 1 7 1 1 9 7 u u +C= cot x cot x+C 9 7 9 7. 41. Use Equation 2(b): 1 1 sin 5xsin 2xdx = 2 cos (5x2x)cos (5x+2x) dx= 2 ( cos 3xcos 7x ) dx 1 1 = sin 3x sin 7x+C 6 14 42. Use Equation 2(a): 1 1 sin 3xcos xdx = 2 sin (3x+x)+sin (3xx) dx= 2 ( sin 4x+sin 2x ) dx 1 1 = cos 4x cos 2x+C 8 4 43. Use Equation 2(c): 1. cos 7 cos 5 d = 2 =. 1 ( cos 2 +cos 12 ) d 2 1 1 +C= sin 2 + sin 12 +C 4 24. cos (7 5 )+cos (7 +5 ) d =. 1 2. 1 1 sin 2 + sin 12 2 12. 44. cos x+sin x 1 cos x+sin x 1 sin 2x dx = 2 sin xcos x dx= 2 (csc x+sec x)dx 1 = ( ln csc xcot x +ln sec x+tan x ) +C [ by Exercise 39 and (1)] 2 45. . 2. 1tan x 2. sec x. (. ). dx= cos xsin x dx= cos 2xdx= 2. 2. 1 sin 2x+C 2. 46. 1. dx cos x1 = cos x1 . (. cos x+1 cos x+1 cos x+1 dx= dx= dx 2 2 cos x+1 cos x1 sin x. ). = cot xcsc xcsc x dx=csc x+cot x+C 2. 2. 2 2. 47. Let u=tan (t ) du=2tsec (t )dt . Then 9.
(39) Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.2 Trigonometric Integrals. tsec. 2. ( t2) tan 4 ( t2) dt= u4. 1 1 5 1 5 2 du = u +C= tan (t )+C . 2 10 10. 7. 6. 2. 48. Let u=tan x , dv=sec xtan xdx du=7tan xsec xdx , v=sec x . Then. tan. 8. xsec xdx = tan x sec xtan xdx=tan xsec x 7tan xsec xsec xdx 7. 7. (. 6. 2. ). =tan xsec x7 tan x tan x+1 sec xdx 7. 6. 2. =tan xsec x7 tan xsec xdx7 tan xsec xdx . 7. 8. 6. /4. Thus, 8 tan xsec xdx=tan xsec x7 tan xsec xdx and 8. dx=. 7. 1 7 tan xsec x 8. /4 0. . 6. 0. 8. tan xsec x. 2 7 7 /4 6 tan xsec xdx= I. 8 8 8 0. 49. Let u=cos x du=sin xdx . Then. sin. 5. ( 1cos. )2 ). (. 2 2. ). x sin xdx= 1u ( du ) 1 5 2 3 2 4 = 1+2u u du= u + u u+C 5 3 1 2 5 3 = cos x+ cos xcos x+C 5 3 Notice that F is increasing when f ( x ) >0 , so the graphs serve as a check on our work. xdx =. 2. (. 50.. sin. 4. xcos xdx = 4. 1 sin 2x 2. 4. 1 1 4 dx= sin 2xdx= 16 16. 1 ( 1cos 4x ) 2. 2. dx. 1 2 12cos 4x+cos 4x dx 64 1 1 1 = x sin 4x + ( 1+cos 8x ) dx 64 2 128 =. (. ). 10.
(40) Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.2 Trigonometric Integrals. 1 1 1 1 x sin 4x + x+ sin 8x +C 64 2 128 8 3 1 1 = x sin 4x+ sin 8x+C 128 128 1024. =. Notice that f (x)=0 whenever F has a horizontal tangent.. 51. 1. sin 3xsin 6xdx = 2. cos (3x6x)cos (3x+6x) dx. 1 (cos 3xcos 9x)dx 2 1 1 = sin 3x sin 9x+C 6 18 =. Notice that f (x)=0 whenever F has a horizontal tangent.. 52.. sec. 4. x 2 x 2 x dx = tan +1 sec dx 2 2 2 x 1 2 2 x , du= sec = u +1 2du [ u=tan dx ] 2 2 2 2 3 2 x 3 x = u +2u+C= tan +2tan +C 3 2 3 2. (. ). Notice that F is increasing and f is positive on the intervals on which they are defined. Also, F has 11.
(41) Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.2 Trigonometric Integrals. no horizontal tangent and f is never zero.. 53. f. ave. 1 2 1 = 2 =0. =. . sin 0u. 0 2. 2. 3. xcos xdx=. 1 2. . sin. 2. (. 2. ). x 1sin x cos xdx. ( 1u2) du [ where u=sin x ]. 1 2 1 2 u +C= cos x+C . 1 2 2 1 2 1 2 (b) Let u=sin x . Then du=cos xdx sin xcos xdx= udu= u +C= sin x+C . 2 2 2 1 1 (c) sin xcos xdx= sin 2xdx= cos 2x+C 3 2 4 (d) Let u=sin x , dv=cos xdx . Then du=cos xdx , v=sin x , 1 2 2 so sin xcos xdx=sin x sin xcos xdx , by Equation .1.2, so sin xcos xdx= sin x+C . 4 2. 54. (a) Let u=cos x . Then du=sin xdx sin xcos xdx= u(du)=. The answers differ from one another by constants. Since 1 1 1 1 1 2 2 2 2 cos 2x=12sin x=2cos x1 , we find that cos 2x= sin x = cos x+ . 4 2 4 2 4 55. For 0<x< /2. 0. 3 , we have 0<sin x<1 , so sin x<sin x . Hence the area is 2. ( sin xsin 3x) dx=0 /2sin x ( 1sin 2x) dx=0 /2cos 2xsin xdx . Now let u=cos x du=sin xdx .. Then area = u ( du ) = u du= 1 0 0 2. 1 2. 1 3 u 3. 1. =. 0. 1 . 3. 1 2 56. sin x>0 for 0<x< , so the sign of 2sin xsin x [ which equals 2sin x sin x ] is the same 2 2 1 2 as that of sin x . Thus 2sin xsin x is positive on and negative on 0, . The , 2 6 2 6 desired area is 12.
(42) Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.2 Trigonometric Integrals. /6. 0 =. ( sin x2sin 2x) dx+ /2/6 ( 2sin 2xsin x) dx. /6. /2. 0 ( sin x1+cos 2x ) dx+ /6 ( 1cos 2xsin x ) dx 1 sin 2x 2. /6. =. cos xx+. =. 3 3 + ( 1 ) + 2 4 6 2. = 1+. 3 2 6. 0. + x. 1 sin 2x+cos x 2. /2 /6. 3 3 + 4 2 6. 57.. It seems from the graph that cos xdx=0 , since the area below the x axis and above the graph 2. 3. 0. looks about equal to the area above the axis and below the graph. By Example 1, the integral is 2 1 3 sin x sin x =0 . Note that due to symmetry, the integral of any odd power of sin x or cos x 3 0 between limits which differ by 2n ( n any integer) is 0 . 58.. It seems from the graph that sin 2 xcos 5 xdx=0 , since each bulge above the x axis seems to 2. 0. have a corresponding depression below the x axis. To evaluate the integral, we use a trigonometric identity: 1 1 2 0sin 2 xcos 5 xdx = 2 0 sin (2 x5 x)+sin (2 x+5 x) dx 1 2 = sin (3 x)+sin 7 x dx 2 0 13.
(43) Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.2 Trigonometric Integrals. . 59. V =. . sin xdx= 2. /2. 2. 1 1 cos (3 x) cos 7 x 3 7 1 1 (11) (11) =0 3 7. 1 2 1 = 2 =. /2. 1 (1cos 2x)dx= 2. 0. 2. . 1 1 x sin 2x 2 4. 0 +0 = 4 2 4. =. /2. 60. Volume = /4 tan 2 x 2dx= /4tan 2 x sec 2 x1 dx= /4tan 2 xsec 2 xdx /4tan 2 xdx. (. 0. ). /4 2. /4. 0. 0. = . u du . 1 3 u 3. =. /4. (. 0. ). 0. 0. ( sec 2x1) dx. tan xx. x=0. /4. 1 3 tan xtan x+x 3. =. 0. /4 0. 1 1+ 3 4. =. =. 2 4 3. 61. Volume = /2 ( 1+cos x ) 212 dx= /2 2cos x+cos 2 x dx 0 0. (. 1 1 2sin x+ x+ sin 2x 2 4. =. /2. =. 0. ). 2+ 4. 2. =2 + 4. 62. Volume = /2 12 ( 1cos x ) 2 dx= /2 2cos xcos 2 x dx 0. 1 1 2sin x x sin 2x 2 4. =. (. 0. t. 2. /2. =. 0. ). 2 0 0 =2 4 4. 63. s= f (t)= sin ucos udu . Let y=cos u dy= sin udu . Then 2. 0. s=. cos t 2. 1 . 1. y dy=. 1 3 y 3. 1 . cos t. =. 1. 1 3. ( 1cos 3 t ) .. 64. (a) We want to calculate the square root of the average value of 2. 2. 2. 2. E(t) = 155sin (120 t) =155 sin (120 t) . First, we calculate the average value itself, by 1 2 integrating E(t) over one cycle (between t=0 and t= , since there are 60 cycles per second) and 60 14.
(44) Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.2 Trigonometric Integrals. 1 0 : 60 1 1/60 2 2 2 1/60 1 = 155 sin (120 t) dt=60 155 1cos (240 t) dt 0 2 1/60 0. dividing by E(t). 2 ave. = 60 1552. 1 2. 1 t sin (240 t) 240. 1/60 0. 2. =60 155. The RMS value is just the square root of this quantity, which is E(t). (b) 220=. 2 220 = E(t) 2. = 30A 2. Thus, 220 =. 2. 2 ave. =. ave. t. 1 2. 1 0 ( 00 ) 60. 2. 155 = 2. 155 110 V. 2. . 1 1/60 2 2 2 1/60 1 A sin (120 t)dt=60A [1cos (240 t)]dt 0 2 1/60 0 1 sin (240 t) 240. 1/60 0. 2. =30A. 1 0 ( 00 ) 60. =. 1 2 A 2. 1 2 A A=220 2 311 V. 2. 65. Just note that the integrand is odd . Or: If m n , calculate 1 sin mxcos nxdx = 2 sin (mn)x+sin (m+n)x dx cos (mn)x cos (m+n)x 1 mn m+n 2 If m=n , then the first term in each set of brackets is zero. =. . . =0. . 1 [cos (mn)xcos (m+n)x]dx . If m n , this is equal to 2 sin (mn)x sin (m+n)x 1 =0 . If m=n , we get mn m+n 2 1 1 sin (m+n)x 2 [1cos (m+n)x]dx= 2 x 2(m+n) = 0= .. 66. sin mxsin nxdx=. . . 1 cos (mn)x+cos (m+n)x dx . If m n , this is equal to 2 sin (mn)x sin (m+n)x 1 + =0 . If m=n , we get mn m+n 2 1 1 sin (m+n)x 2 [1+cos (m+n)x]dx= 2 x + 2(m+n) = +0= .. 67. cos mxcos nxdx=. 15.
(45) Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.2 Trigonometric Integrals. 1 68. . . 1 f (x)sin mxdx= . . . m. n=1ansin nx. sin mx dx=. m n=1. a. n . . sin mxsin nxdx . By. a Exercise 66 , every term is zero except the m th one, and that term is. m. . =a . m. 16.
(46) Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.3 Trigonometric Substitution. 1. Let x=3sec , where 0 <. 3 or < . Then dx=3sec tan d and 2 2. (. ). 2 2 2 2 9sec 9 = 9 sec 1 = 9tan x 9 = = 3 tan =3tan for the relevant values of . . 1 2. x. 2. dx=. x 9. 1 2. 9sec 3tan . 1 1 1 3sec tan d = cos d = sin +C= 9 9 9. Note that sec ( + )=sec , so the figure is sufficient for the case < 2. Let x=3sin , where . 2. x 9 +C x. 3 . 2. . Then dx=3cos d and 2 2. (. ). 2 2 2 2 99sin = 9 1sin = 9cos 9x = = 3 cos =3cos for the relevant values of \italic{\theta}\,\,.. x. 3. 2 9x dx =. 3 sin 3. 3cos 3cos d =3 sin cos d. 3. 5. 3. (. 2. ). = 35 sin 2 cos 2 sin d =35 1cos 2 cos 2 sin d. ( (. ). = 35 1u2 u2 (du)[u=cos ,du=sin d ] 1 5 1 3 1 1 5 4 2 5 5 5 3 = 3 u u du=3 u u +C=3 cos cos 5 5 3 3 5. = 3. ). 1 5. ( 9x2) 5/2 1 ( 9x2) 3/2 5. 3. 3. 3. +C. +C. 3. 1.
(47) Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.3 Trigonometric Substitution. 1 2 9x 5. (. =. 3. Let x=3tan , where . ) 5/23 ( 9x2) 3/2+Cor 15 ( x2+6) ( 9x ) 3/2+C. 2 < < . Then dx=3sec d and 2 2. (. ). 2 2 2 2 9tan +9 = 9 tan +1 = 9sec x +9 = = 3 sec =3sec for the relevant values of .. . x. 3. 2. dx =. x +9. . 3. 3. 3 tan 2 3 3 3 2 3sec d =3 tan sec d =3 tan tan sec d 3sec . (. ). (. ). = 33 sec 2 1 tan sec d =33 u21 du[u=sec ,du=sec tan d ] 1 3 3 u u +C=3 3. 3. = 3 =. 1 2 x +9 3. (. ) 3/29. 2. 1 3 sec sec 3. x +9 +C or. 1 2 x 18 3. (. 3. +C=3. ). 1 3. ( x2+9) 3/2 3. 3. 2. x +9 3. +C. 2. x +9 +C. 4. Let x=4sin , where /2 /2 . Then dx=4cos d and 2. 2. 2. 16x = 1616sin = 16cos =4 cos =4cos . When x=0 , 4sin =0 =0 , and when 3 . Thus, substitution gives = x=2 3 , 4sin =2 3 sin = 2 3. 2.
(48) Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.3 Trigonometric Substitution. 2 3. /3. . x. =. 3. 0. dx. /3 3. 2. 16x. 0. 3. . =4. /3. 3. 4 sin 3 3 4cos d =4 sin d 4cos 0. ( 1cos 2 ) sin d. 0 1/2 3. 1 3 1/2 u u 1 3 1 1 =64 3. 2 ( 1u ) du=64. =4. 1. 1 1 2 24. =64. 2. . 5 24. =. 40 3. =. 3 1 + 8 4 24. 2. Or: Let u=16x , x =16u , du=2xdx . 5. Let t=sec , so dt=sec tan d , t= 2 = 2. 2. /3. 1 t. 3. 2. /3. . dt =. 1 3. sec tan . /4. t 1. sec tan d =. 1 1 (1+cos 2 )d = 2 2 /4. . 1 + 3 2. 1 2. =. . 3 2. +. /3. 1 2. sec . /4. /3. =. , and t=2 = . Then 4 3. /4. /4. =. 3 1 + 4 2 12. 1 2. 2. 6. Let x=2tan , so dx=2sec d , x=0 =0 , and x=2 =. 0x. 2 3. /4 3. 5 /4. x +4 dx = 0 2 tan 2sec 2sec d =2 2. 3. 5 /4. =2. 0. 2. 2. 2. (sec 1)sec sec tan d (u 1)u du [u=sec ,du=sec tan d ]. 1. 2. (u u )du=2. =32. . Then 4. tan sec sec tan d. 2. 5. =2. 2. 0. 1. 5. =2. 2. 2. /3. 1 sin 2 2. 1 + 1 4 2. . d = cos d. 2. 4. 2. 2. 2 2 2+ 15 15. 5. 1 5 1 3 u u 5 3 64 = ( 2 +1 ) 15. 2 1. 5. =2. 1 1 4 2 2 2 5 3. . 1 1 5 3. 3.
(49) Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.3 Trigonometric Substitution. 2. 2. Or: Let u=x +4 , x =u4 , du=2xdx . 7. Let x=5sin , so dx=5cos d . Then 1 1 dx = 5cos d 2 2 2 2 sin 5cos 5 x 25x 1 1 2 = csc d = cot +C 25 25 2. 25x +C x. = 1 25. 8. Let x=asec , where 0 <. 3 or < . Then dx=asec tan d and 2 2. 2. 2. x a =atan ,. so. . 2. 2. x a x. 4. dx =. atan . . 4. 4. a sec . 1 =. =. sin 2. asec tan d. 2. cos d. a. 1. 2 2 3/2 ( x a ) sin +C= +C 3. 2. 3a. 9. Let x=4tan , where . 2 3. 3a x. 2 < < . Then dx=4sec d and 2 2 4.
(50) Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.3 Trigonometric Substitution. 2. 2. 2. x +16 = 16tan +16 = 16(tan +1) 2. = 16sec =4 sec =4sec for the relevant values of .. dx. . 2. 4sec d = = sec d =ln sec +tan +C 2 1 4sec x +16 2. x +16 x + 4 4. =ln =ln. (. +C =ln 1. ). 2. 2. x +16 +x ln 4 +C. 1. x +16 +x +C , where C=C ln 4 . 1. 2. x +16 +x>0 , we don’t need the absolute value.). (Since. 10. Let t= 2 tan , where 2. 2 < < . Then dt= 2 sec d and 2 2. 2. 2. 2. t +2 = 2tan +2 = 2(tan +1) = 2sec = 2 sec = 2 sec for the relevant values of .. . t. 5. 2. t +2. 5. dt =. 4 2 tan 2 sec . (. )2. (. )2. 2 sec d =4 2 tan sec d =4 2 sec 1 sec tan d 2. 5. (. 2. ). =4 2 u 1 du [ u=sec , du=sec tan d ] =4 2 u 2u +1 du 2. 4. 2. 5.
(51) Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.3 Trigonometric Substitution. 4 2 1 5 2 3 4 2 u u +u +C= u(3u 10u +15)+C 15 5 3. =4 2. 2. t +2. 4 2 = 15 4 15 1 = 15. 2 2. t +2 . =. 2. 3. 1. 0. 2. 2. 2. t +2 10 +15 2. +C. 1 4 2 2 3(t +4t +4)20(t +2)+60 +C 4 4. 2. 1 1 . Then x= sin , dx= cos d , and 2 2 2 2. 2. 14x = 1 ( 2x ) =cos . 1 2 14x dx = cos 2 cos 1 1 = + sin 2 4 2 1 1 = sin (2x)+2x 4. 12. x. 2. t +2 (3t 8t +32)+C. 11. Let 2x=sin , where 2. 2. (t +2). x +4 dx= 2. 5 4. u. 1 du 2. 13. Let x=3sec , where 0 <. 1 ( 1+cos 2 ) d 4 1 +C= ( +sin cos )+C 4 d =. 2. 14x. +C. 2. [ u=x +4 , du=2xdx ] =. 1 2 3/2 5 1 u = ( 5 5 8 ) 4 3 2 3. 3 or < . Then dx=3sec tan d and 2 2. 2. x 9 =3tan ,. so. . 2. x 9 x. 3. 3tan . 2. 1 tan 3sec tan d = d dx = 3 2 3 27sec sec . 6.
(52) Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.3 Trigonometric Substitution. =. 1 1 1 1 1 1 1 2 sin d = (1cos 2 )d = sin 2 +C= sin cos +C 3 3 2 6 12 6 6 x 3. 1 1 = sec 6. 2. x 9 3 1 1 +C= sec x x 6. 1 6. x 3. 2. . x 9 2. +C. 2x. 14. Let u= 5 sin , so du= 5 cos d . Then 1 1 du = 5 cos d = csc d 2 5 sin 5 cos 5 u 5u 1 ln csc cot +C [byExercise2.39 = 5 =. 1 ln 5. 5 u. =. 1 ln 5. 5 5u u. 15. Let x=asin , where 2. x dx. ( a2x2) 3/2. 2. 5u u. . =. ( sec. 2. 3. 3. a cos 2. +C. 2. +C. . Then dx=acos d and 2 2. a sin acos d. =. 2. = tan d 2. ). 1 d =tan +C. 7.
(53) Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.3 Trigonometric Substitution. x =. 2. 2. sin. 1. a x. x +C a. 16. Let 4x=3sec , where 0 <. 3 3 or < . Then dx= sec tan d and 2 2 4. 2. . 16x 9 =3tan , so 3 sec tan d dx 4 = 2 2 3 2 2 x 16x 9 sec 3tan 4. 2. 2. 16x 9 +C= 4x. 4 4 4 = cos d = sin +C= 9 9 9. 17. Let u=x 7 , so du=2xdx . Then 2. x 2. 16x 9 +C 9x. dx=. x 7 2. 2. 2. 2. 2. 1 1 1 2 du= 2 u +C= x 7 +C . 2 2 u 2. 2. 2. 2. 18. Let ax=bsec , so ( ax ) =b sec ( ax ) b =b sec b =b b 2 2 ax b =btan , dx= sec tan d , and a. ( sec 2 1) =b2tan 2. . So. ( ). 8.
(54) Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.3 Trigonometric Substitution. . dx 2. 2 3/2. ( ax ) b. =. . =. b sec tan a 3. b tan 1 2. 1. 2. ab. csc +C=. 1 2. sec 2. tan ax 2. ab. ab. 19. Let x=tan , where . . 3. d =. d =. 1. 2. ab +C=. 2. 2. ( ax ) b. b. 2 < < . Then dx=sec d and 2 2. cos 2. sin x 2. 1. d =. 2. ab. csc . cot d. +C 2. ( ax ) b 2. 1+x =sec , so. 2 sec sec 2 2 1+x = sec d = (1+tan )d dx tan tan x. = (csc +sec tan )d =ln csc cot +sec +C [ by Exercise 8.2.39] 2. =ln. 1+x 1 x x. 20. Let t=5sin , where . . t 25t. 2. 2. 1+x +C=ln 1. +. 2. 1+x 1 x. . Then dt=5cos d and 2 2. 2. + 1+x +C. 2. 25t =5cos , so. dt = 5sin 5cos 5cos d =5 sin d 2. =5cos +C=5. 25t 2 +C= 25t +C 5. 2. Or: Let u=25t , so du=2t dt .. 9.
(55) Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.3 Trigonometric Substitution. 2. 1 ( 4u ) and 9 1 1 4 1/2 3/2 du= 4u u du 18 162 0 2. 21. Let u=49x du=18xdx . Then x =. 0. 2/3 3. x. 1. 4 9 (4u)u 0. 2. 49x dx =. Or: Let 3x=2sin , where 22. Let x=tan , where =. 0 1. , so 4 2. x +1 dx =. =. 0. 1 162. (. ). 64 64 3 5. =. 64 1215. . 2 2. 2 < < . Then dx=sec d , 2 2. /4. 0. 4. 8 3/2 2 5/2 u u 3 5. 1 162. =. 1/2. /4. sec sec d = 2. 0. 2. x +1 =sec and x=0 =0 , x=1. 3. sec d. 1 /4 sec tan +ln sec +tan [by Example 8.2.8] 0 2 1 1 = 2 1+ln ( 1+ 2 ) 0ln (1+0) = 2 +ln ( 1+ 2 ) 2 2. =. 2. 2. 2. 23. 5+4xx =(x 4x+4)+9=(x2) +9 . Let x2=3sin , . . , so dx=3cos d . Then 2 2. 5+4xx dx = 9(x2) dx= 99sin 3cos d 2. 2. 2. = 9cos 3cos d = 9cos d 2. 2. 10.
(56) Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.3 Trigonometric Substitution. 9 9 1 (1+cos 2 )d = + sin 2 +C 2 2 2 9 9 9 9 = + sin 2 +C= + (2sin cos )+C 2 4 2 4. =. (2. 2. 2. x2 3 x2 3. 9 1 = sin 2 9 1 = sin 2. ). 5+4xx 9 x2 + +C 3 2 3 1 2 + (x2) 5+4xx +C 2. 2. 2. 2. 24. t 6t+13= t 6t+9 +4= ( t3) +2 . Let t3=2tan , so dt=2sec d . Then. . dt 2. t 6t+13. =. 1. . 2. 2. ( 2tan ) +2 =. 2. 2sec 2sec d = d 2sec 2. sec d =ln. sec +tan +C. 1. [by Formula 8.2.1]. 2. t 6t+13 t3 + 2 2. = ln. +C. 1. 2. t 6t+13 +t3 +CwhereC=C ln 2. = ln. 1. 25. 9x +6x8= ( 3x+1 ) 9 , so let u=3x+1 , du=3dx . Then 2. u=3sec , where. 2. dx 2. 9x +6x8. =. 1 du 3 2. . Now let. u 9 11.
(57) Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.3 Trigonometric Substitution. 0 <. . 3 or < . Then du=3sec tan d and 2 2. 1 du 3. =. 2. u 9. . (. sec tan d 1 1 1 = sec d = ln sec +tan +C = ln 1 3 3tan 3 3. 2. u+ u 9 3. +C. 1. 1 1 2 2 ln u+ u 9 +C= ln 3x+1+ 9x +6x8 +C 3 3. = 2. 2. u 9 =3tan , so. 2. ). 2. 2. ( 2sin +2 ). 26. 4xx = x 4x+4 +4=4 ( x2 ) , so let u=x2 . Then x=u+2 and dx=du , so. . 2. x dx 2. =. ( u+2 ) du. 4xx. 4u. (. 2. =. 2. 2cos . 2cos d. ). =4 sin +2sin +1 d =2 ( 1cos 2 ) d +8 sin d +4 d =2 sin 2 8cos +4 +C =6 8cos 2sin cos +C 1 1 2 1 2 =6sin u 4 4u u 4u +C 2 2 x2 x2 1 2 2 =6sin 4 4xx 4xx +C 2 2. 2. 2. 2. 27. x +2x+2=(x+1) +1 . Let u=x+1 , du=dx . Then. . dx. ( x2+2x+2) 2. = =. . du. ( u2+1) 2. cos. 2. =. d =. 2. sec d 4. sec . 2. where u=tan ,du=sec d , 2. 2. and u +1=sec x. 1 1 ( 1+cos 2 ) d = ( +sin cos )+C 2 2. 12.
(58) Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.3 Trigonometric Substitution. 1. 1 = 2. (. 2. u. tan u+. 1+u. ). 2. +C=. 2. 1 2. x+1. 1. tan (x+1)+. +C. 2. x +2x+2. 2. 28. 54xx = x +4x+4 +9=9 ( x+2 ) . Let u=x+2 du=dx . Then dx where u=3sin , du=3cos d , du 3cos d = = 5 2 2 5/2 2 5/2 (3cos ) and 9u =3cos 54xx 9u 1 1 1 1 4 2 2 3 = sec d = tan +1 sec d = tan +tan 3 81 81 81. (. ). (. ). (. u. 1 = 243. 3. 2 3/2. ( 9u ). +. ). 3u 9u. 2. 1 +C= 243. 3. (x+2). 2 3/2. ( 54xx ). +. +C 3(x+2). +C 2. 54xx. 2. 29. Let u=x , du=2xdx . Then. x. 1x dx = 1u 4. whereu=sin ,du=cos d ,. 1 1 du = cos cos d 2 2. 2. 2. and 1u =cos . 1 1 1 1 1 1 (1+cos 2 )d = + sin 2 +C= + sin cos +C 2 2 4 8 4 4 1 1 1 1 2 1 2 1 2 4 = sin u+ u 1u +C= sin (x )+ x 1x +C 4 4 4 4. =. 30. Let u=sin t , du=cos tdt . Then /2. 1. . cos t 2. . dt =. 1+sin t. 0. 1+u. 0. 2. 2. x +a. . 0. 0. 0. sec d = ln sec +tan . (. 2 +1 ) ln (1+0)=ln. 31. (a) Let x=atan , where =. dt= . /4. =ln. dx. 2. 2. 2. where u=tan , du=sec d ,. 1 2 sec d sec . /4. =. . /4. 1. (. 2. and 1+u =sec [ by (1) in Section 8.2]. 2 +1 ). . Then < < 2 2. 2. 2. x +a =asec and. asec d = sec d =ln sec +tan +C =ln 1 asec . 2. 2. x +a x + a a. +C. 1 13.
(59) Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.3 Trigonometric Substitution. (. 2. 2. = ln x+ x +a. ) +C whereC=C ln 1. 2. a. 2. (b) Let x=asinh t , so that dx=acosh t dt and x +a =acosh t . Then acosh t dt 1 x dx = acosh t =t+C=sinh a +C . 2 2 x +a 32. (a) Let x=atan , I = =. 2. x. . 2 3/2. ( x +a ) 2. ( sec . dx=. . Then < < 2 2 2. 2. 3. 3. a tan . 2. 2. tan sec 1 asec d = d = d sec sec 2. a sec . cos ) d =ln sec +tan sin +C 2. 2. x +a x + a a. = ln. . x. (. 2. 2. +C=ln x+ x +a. 2. 2. ). x +a. x 2. +C 2. 1. x +a. (b) Let x=asinh t . Then I =. . 2. 2. 3. 3. a sinh t. a cosh t 1 x = sinh a . (. ). acosh t dt= tanh t dt= 1sech t dt=ttanh t+C 2. x 2. 2. +C 2. a +x. 2. 33. The average value of f (x)= x 1 /x on the interval 1,7 is 7. 1 71. 1. 2. 1 x 1 dx = 6 x. . 0. tan sec tan d sec . where x=sec ,dx=sec tan d , 2. 1. x 1 =tan ,and =sec 7. 1 1 2 2 tan d = (sec 1)d 6 0 6 0 1 1 = tan = (tan ) 0 6 6 1 1 = 48 sec 7 6 =. (. ). 14.
(60) Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.3 Trigonometric Substitution. 2. 2. 34. 9x 4y =36 y= area = 2. 3 2. 3 2. 3 2. 2. x 4 . x 4 dx=3 2. 3 2. 2. x 4 dx. . = 3 2tan 2sec tan d 0. where x=2sec , dx=2sec tan d , 1 3 =sec 2. . = 12 = 12. 2 3 ( sec 1 ) sec d =12 ( sec sec ) d 0 0. 1 ( sec tan +ln sec +tan ) ln sec +tan 2. 3 5 ln 4. 5 3 + 2 2. 0. . = 6 sec tan ln sec +tan =6. . 0. =. 9 5 6ln 2. 3+ 5 2. r 1 1 2 (rcos )(rsin )= r sin cos . Area of region PQR= rcos 2 2 . Then we obtain x=rcos u dx=rsin udu for u 2 1 2 2 2 2 2 = rsin u (rsin u)du=r sin udu= r (usin ucos u)+C r x dx 2 1 2 1 1 2 2 = r cos (x/r)+ x r x +C 2 2 so. 35. Area of POQ=. area of region = 1 r 2cos 1(x/r)+x r 2x2 2 1 2 = 0 r +rcos rsin 2. (. 2. 2. r x dx . Let. r rcos . ). 15.
(61) Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.3 Trigonometric Substitution. 1 2 1 2 r r sin cos 2 2. =. and thus, ( area of sector POR ) = ( area of POQ ) + ( area of region PQR ) = 36. Let x= 2 sec , where 0 <. . dx. 1 2 r . 2. 3 or < , so dx= 2 sec tan d . Then 2 2. 2 sec tan d. =. 4. 4 2 4sec 2 tan x x 2 1 1 3 2 = cos d = 1sin cos d 4 4 1 1 3 = sin sin +C [substitute u=sin ] 4 3. (. 1 = 4. 2. (. ). ) 3/2. 2. x 2 x 2 3 x 3x. +C. From the graph, it appears that our answer is reasonable. 2. 37. From the graph, it appears that the curve y=x 2. 2. 4x and the line y=2x intersect at about x=0.81. 2. 4x >2x on ( 0.81,2 ) . So the area bounded by the curve and the line is 2 2 1 2 2 2 2 2 2 . To evaluate the integral, we put A x 4x (2x) dx= x 4x dx 2x x 0.81 0.81 0.81 2 . Then x=2sin , where 2 2. and x=2 , with x. 16.
(62) Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.3 Trigonometric Substitution. 1. dx=2cos d , x=2 =sin 1=. 0.81x 2. 2. /2. 0.417 4sin. 2. 4x dx. 2 2. /2. (2cos )(2cos d )=4. 2. /2. 1 sin 4 4. = 2 Thus, A 2.81. 1 , and x=0.81 =sin 0.405 0.417 . So 2. 0.417. 1 1 2 2 2 2 0.81 0.81 2 2 2. 2. 1 0 0.417 (0.995) 2 4. =2. 2. 38. Let x=btan , so that dx=bsec d and. 0.417. /2. sin 2 d =4. 1 (1cos 4 )d 0.417 2. 2.81. 2.10 .. 2. x +b =bsec .. La. . E ( P) =. a. b 4 . 2 2 3/2 ( x +b ) 0. b dx= 4 . 2. . 2. 1. . 2. ( bsec ). 0 1. bsec d. 3. 2. 1 = sec d = cos d = sin 4 b 4 b 4 b 0. = 4 b 0. 0. 1. La. x 2. = 2. x +b. a. 0. 1. 4 b. La 2. 0. 2. + 2. (La) +b. 2. . 2. . 1. a 2. 2. a +b. 2. 39. Let the equation of the large circle be x +y =R . Then the equation of the small circle is 2. 2. 2. 2. 2. x +(yb) =r , where b= R r is the distance between the centers of the circles. The desired area is 17.
(63) Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.3 Trigonometric Substitution. A=. r. r. ( b+. 2. 2. r x. r. r. 0. 0. = 2 bdx+2. ). 2. r x dx2 2. dx=2. 2. R x. 2. r. 2. 0. ( b+. 2. 2. 2. 2. r x R x. ) dx. 2. R x dx. 0. 2. 2. R r . To evaluate the other two integrals, note that 1 2 2 2 2 2 a x dx = a cos d [x=asin ,dx=acos d ] = 2 a ( 1+cos 2 ) d 1 2 1 1 2 = a + sin 2 +C= a ( +sin cos )+C 2 2 2. The first integral is just 2br=2r. . r. 2. = a arcsin 2 so the desired area is 2. A = 2r. 2. 2. x a. 2. 2. 2. R r + r arcsin(x/r)+x r x 2. 2. R r +r. = 2r. 2. 2. 2. 2. 2. a x a +C= arcsin a 2. x a. a + 2. r. 2. R arcsin(x/R)+x. 0. 2. 2. R arcsin(r/R)+r. R r. 2. =r. 2. 2. x a 2. R x 2. R r +. +. x 2. 2. 2. a x +C. r 0. 2 2 r R arcsin(r/R) 2. 40. Note that the circular crosssections of the tank are the same everywhere, so the percentage of the total capacity that is being used is equal to the percentage of any crosssection that is under water. The underwater area is A = 2. 2 5. 2. 25y dy. = 25 arcsin(y/5)+y = 25 arcsin. 2. 25y. 2 5. y=5sin ]. 25 2 2 +2 21 + 58.72ft 2 5. so the fraction of the total capacity in use is. A 2. (5). 58.72. 0.748 or 74.8% . 25. 18.
(64) Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.3 Trigonometric Substitution. 2. 2. 2. 2. 2. 41. We use cylindrical shells and assume that R>r . x =r (yR) x= r (yR) , so 2. 2. g(y)=2 r (yR) and V =. R+r. Rr 2 y 2 r. r. r (yR) dy= 4 (u+R) r u du 2. 2. 2. r. = 4 = 2 Rr. . 1 2 2 r u 3. (. 2 /2. 2. r. = 4 u r u du+4 R 2. 2. ) 3/2. /2(1+cos 2. r r. r r. 2. 2. where u=rsin ,du=rcos d in the second tegral. r u du /2 2. +4 R. 2. r cos d =. /2. )d =2 Rr. 2. +. [where u=yR]. 1 sin 2 2. 4 2 /2 2 (00)+4 Rr cos d /2 3 /2. 2. /2. Another method: Use washers instead of shells, so V =8 R evaluate the integral using y=rsin .. 2. =2 Rr. r 0. 2. 2. r y dy as in Exercise 6.2.61(a) , but. 19.
(65) Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.4 Integration of Rational Functions by Partial Fractions. 2x A B = + (x+3)(3x+1) x+3 3x+1 1 1 A B 1 C (b) 3 2 = = = + + 2 2 2 x x+1 x +2x +x x(x +2x+1) x(x+1) (x+1) 1. (a). A B C + + 2 3 2 2 x+1 x x (x+1) x x +x x1 x1 A Bx+C (b) 3 = = + 2 2 x x +x x x +1 x +1 x1. 2. (a). x1. =. (. ). 2. 3. (a). =. =. 2. x +3x4. 2 A B = + (x+4)(x1) x+4 x1 2. x. 2. (b) x +x+1 is irreducible, so. (. 2. (x1) x +x+1. 4. (a). x. 3. 13x+12. =x4+. 2. 2. x +4x+3 2x+1. =x4+. =. ). A Bx+C . + 2 x1 x +x+1. 13x+12 A B =x4+ + (x+1)(x+3) x+1 x+3. x +4x+3 A Dx+E Fx+G B C (b) = + + + 2 + 3 2 2 x+1 2 3 2 2 x +4 (x +4) (x+1) (x+1) (x+1) (x +4) 5. (a) x. 4. 4. 4. x 1. =. (x 1)+1 4. x 1. =1+. 1. [ or use long division] =1+. 4. x 1. 1. =1+. =1+. 2. (x1)(x+1)(x +1) 4. (b). 2. t +t +1. ( t2+1) ( t2+4) 2. 6. (a). x 3. =. At+B 2. +. t +1. 2. (x +x)(x x+3). =. Ct+D 2. x 2. 2. (x 1)(x +1). A B Cx+D + + 2 x1 x+1 x +1. t +4. 4. 1 2. +. Et+F. ( t2+4) 2. 4 2. x(x +1)(x x+3). =. x 2. 3 2. (x +1)(x x+3). =. Ax+B 2. x +1. +. Cx+D 2. x x+3. (b) 1.
(66) Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.4 Integration of Rational Functions by Partial Fractions. 1 6. x x 7. . 3. 1. = x. 3. =. ( x31). 1 3. (. 2. x (x1) x +x+1. ). x (x6)+6 6 dx= dx= 1+ x6 x6 x6. =. A B C D Ex+F + + + + 2 2 3 x1 x x +x+1 x x dx=x+6ln x6 +C. 8.. . 2. 2. r r 16 16 16 dr = + dr= r4+ r+4 r+4 r+4 r+4 1 2 = r 4r+16ln r+4 +C 2. dr [ or use long division]. x9 A B = + . Multiply both sides by (x+5)(x2) to get x9=A(x2)+B(x+5) . (x+5)(x2) x+5 x2 Substituting 2 for x gives 7=7B B=1 . Substituting 5 for x gives 14=7A A=2 . Thus, x9 2 1 (x+5)(x2) dx= x+5 + x2 dx=2ln x+5 ln x2 +C 9.. A B 1 = + 1=A(t1)+B(t+4) . (t+4)(t1) t+4 t1 1 1 t=1 1=5B B= . t=4 1=5A A= . Thus, 5 5 1 1/5 1/5 1 1 1 (t+4)(t1) dt= t+4 + t1 dt= 5 ln t+4 + 5 ln t1 +C or 5 ln 10.. 11.. 1 2. x 1. =. 2. 3. 1 2. x 1. 1/2 1/2 + x+1 x1. dx = 2. = . 12.. x1 2. +C. 1 A B = + . Multiply both sides by (x+1)(x1) to get 1=A(x1)+B(x+1) . (x+1)(x1) x+1 x1. Substituting 1 for x gives 1=2B B= 3. t1 t+4. =. 1 1 . Substituting 1 for x gives 1=2A A= . Thus, 2 2. dx= . 1 1 ln x+1 + ln x1 2 2. 3 2. 1 1 1 1 1 1 3 ln 4+ ln 2 ln 3+ ln 1 = (ln 2+ln 3ln 4) or ln 2 2 2 2 2 2 2. A B + . Multiply both sides by (x+1)(x+2) to get x1=A(x+2)+B(x+1) . x+1 x+2. x +3x+2 Substituting 2 for x gives 3=B B=3 . Substituting 1 for x gives 2=A . Thus,. 2.
(67) Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.4 Integration of Rational Functions by Partial Fractions. 1. 0. 1. 2 3 + x+1 x+2. dx =. x1 2. 0. x +3x+2. =(2ln 2+3ln 3)(2ln 1+3ln 2)=3ln 35ln 2 ax. 13. . 2. x bx. dx=. 14. If a b ,. 1 1 x+a x+b. dx. 1. (x+a)(x+b) = ba ( ln dx 2. =. (x+a) 15.. 2x+3 2. =. A + x+1. (x+1) to get A=2 . Now 1. 0. orln. 27 32. , so if a b , then. x+a ln x+b ) +C=. 1 ln ba. x+a x+b. +C. 1 +C . x+a. B. ( x+1 ). 2. 2x+3=A(x+1)+B . Take x=1 to get B=1 , and equate coefficients of x. 1. 2x+3 2. dx =. (x+1). . 2 + x+1. 0. = 2ln 2 3. 16.. 0. ax a dx= dx=aln xb +C x(xb) xb. 1 1 = (x+a)(x+b) ba. If a=b , then . 1. dx= 2ln x+1 +3ln x+2. x 4x10 2. =x+1+. 1. ( x+1 ). 2. 1 dx= 2ln (x+1) x+1. 1 0. 1 1 ( 2ln 11 ) =2ln 2+ 2 2. 3x4 3x4 A B . Write . Then 3x4=A(x+2)+B(x3) . = + (x3)(x+2) x3 x+2 (x3)(x+2). x x6 Taking x=3 and x=2 , we get 5=5A A=1 and 10=5B B=2 , so 1. 0. 3. x 4x10 2. x x6. 1. dx = 0. =. x+1+. 1 2 + x3 x+2. dx=. 1 2 x +x+ln x3 +2ln (x+2) 2. 1 0. 1 3 3 3 +1+ln 2+2ln 3 ( 0+0+ln 3+2ln 2 ) = +ln 3ln 2= +ln 2 2 2 2. 3.
(68) Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.4 Integration of Rational Functions by Partial Fractions. 2. 4y 7y12 A B C 2 = + + 4y 7y12=A(y+2)(y3)+By(y3)+Cy(y+2) . Setting y=0 gives 17. y(y+2)(y3) y y+2 y3 9 1 12=6A , so A=2 . Setting y=2 gives 18=10B , so B= . Setting y=3 gives 3=15C , so C= . 5 5 Now 2. 1. 2. 4y 7y12 dy = y(y+2)(y3). 2. 1. 2 9/5 1/5 + + y y+2 y3. 9 1 dy= 2ln y + ln y+2 + ln y3 5 5. 2 1. 1 9 1 9 ln 4+ ln 12ln 1 ln 3 ln 2 5 5 5 5 1 9 27 9 9 9 8 18 = 2ln 2+ ln 2 ln 2 ln 3= ln 2 ln 3= (3ln 2ln 3)= ln 5 5 5 5 5 5 3 5. = 2ln 2+. 2. x +2x1. 18.. 3. x x. 2. x +2x1 A B C = = + + . Multiply both sides by x(x+1)(x1) to get x(x+1)(x1) x x+1 x1. 2. x +2x1=A(x+1)(x1)+Bx(x1)+Cx(x+1) . Substituting 0 for x gives 1= A A=1 . Substituting 1 for x gives 2=2B B=1 . Substituting 1 for x gives 2=2C C=1 . Thus,. . 2. x +2x1. 19.. 1 1 1 + x x+1 x1. dx=. 3. x x 1. =. 2. ( x+5) (x1). A + x+5. B. ( x+5). 2. +. dx=ln x ln x+1 +ln x1 +C=ln. x(x1) x+1. +C .. C 2 1=A(x+5)(x1)+B(x1)+C(x+5) . Setting x=5 gives x1. 1 1 . Setting x=1 gives 1=36C , so C= . Setting x=2 gives 6 36 1 1 3 1 1 2 1=A(3)(3)+B(3)+C 3 =9A3B+9C=9A+ + =9A+ , so 9A= and A= . Now 2 4 4 4 36 1/36 1/36 1 1/6 + dx dx = 2 2 x+5 x1 ( x+5) (x1) ( x+5) 1 1 1 = ln x+5 + + ln x1 +C 36 6(x+5) 36 1=6B , so B=. ( ). 2. 20.. x. 2. (x3)(x+2). =. A B + + x3 x+2. C. 2. 2. 2. x =A(x+2) +B(x3)(x+2)+C(x3) .. (x+2) 4 9 2 Setting x=3 gives A= . Take x=2 to get C= , and equate the coefficients of x to get 1=A+B 5 25 4.
(69) Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.4 Integration of Rational Functions by Partial Fractions. B=. . 16 . Then 25 2. dx = 2 (x3)(x+2). 2. 2. 21.. 3. 2. =. 5x +3x2 2. =. x ( x+2 ). x +2x. dx. 16 4 9 ln x3 + ln x+2 + +C 25 5(x+2) 25. =. 5x +3x2. 9/25 16/25 4/5 + 2 x3 x+2 (x+2). . x. A B C 2 + + . Multiply by x (x+2) to get 2 x+2 x x. 2. 2. 5x +3x2=Ax(x+2)+B(x+2)+Cx . Set x=2 to get C=3 , and take x=0 to get B=1 . Equating the 2. coefficients of x gives 5=A+C A=2 . So. . 2. 5x +3x2 3. 2. x +2x 1. 22.. 2. 2. 2 1 3 + 2 x+2 x x. dx=. =. s (s1). dx=2ln x +. 1 +3ln x+2 +C . x. A B C 2 2 2 2 D + + + 1=As(s1) +B(s1) +Cs (s1)+Ds . Set s=0 , giving B=1 . 2 s1 2 s s (s1) 3. Then set s=1 to get D=1 . Equate the coefficients of s to get 0=A+C or A=C , and finally set s=2 to get 1=2A+14A+4 or A=2 . Now 2 2 1 1 2 ds 2 = s + 12 s1 + 1 2 ds=2ln s s 2ln s1 s1 +C . s (s1) s (s1) 2. x. 23.. 3. =. (x+1). A + x+1. B 2. C. +. (x+1). 3. 2. 2. . Multiply by ( x+1 ) to get x =A(x+1) +B(x+1)+C . Setting. 3. (x+1). 2. x=1 gives C=1 . Equating the coefficients of x gives A=1 , and setting x=0 gives B=2 . Now. . 2. x dx 3. =. (x+1). 1 x+1. 2 2. 1. +. (x+1). 3. (x+1). (x+1)1 1 x = =1 24. , so x+1 x+1 x+1. . x. 3 3. (x+1). dx=. 1. 3 + x+1. dx=ln x+1 +. 3 2. (x+1). x. 3. 1 = 1 3 x+1 (x+1) . 1 3. (x+1). 2 x+1 3. =1. 1 2. +C .. 2(x+1). 3 + x+1. dx=x3ln x+1 . 3 2. (x+1). 3 + x+1. 1. . 3. . Thus,. (x+1). 1 2. +C .. 2(x+1). 5.
(70) Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.4 Integration of Rational Functions by Partial Fractions. 10. 25.. (. 2. (x1) x +9. (. 2. ). ). =. A Bx+C 2 + 2 . Multiply both sides by (x1) x +9 to get x1 x +9. (. ). 10=A x +9 +(Bx+C)(x1) ( * ). Substituting 1 for x gives 10=10A A=1 . Substituting 0 for x gives 2. 10=9AC C=9(1)10=1 . The coefficients of the x terms in ( * ) must be equal, so 0=A+B B=1. Thus, 10 1 1 x1 x 1 dx = + dx= dx 2 2 2 2 x1 x1 (x1)(x +9) x +9 x +9 x +9 x 1 1 2 2 1 [Formula10] +C =ln x1 ln x +9 [let u=x +9] tan 3 2 3. (. 2. 26.. x x+6 3. x +3x. 2. =. x x+6. =. x ( x +3) 2. ). A Bx+C 2 2 2 + 2 . Multiply by x x +3 to get x x+6=A x +3 +(Bx+C)x . x x +3. (. ). (. ). 2. Substituting 0 for x gives 6=3A A=2 . The coefficients of the x terms must be equal, so 1=A+B B=12=1 . The coefficients of the x terms must be equal, so 1=C . Thus,. . 2. x x+6 3. 2 x1 + 2 x x +3. dx =. x +3x. =2ln x 3. 27.. 2. x +x +2x+1. ( x2+1) ( x2+2). 3. 2. 3. 2. 3. 2. =. x 2. . x +3. 1. dx. 2. x +3. 1 1 2 1 x ln x +3 tan +C 2 3 3. (. Ax+B 2. x +1. (. 2 x. dx=. 2. +. ). Cx+D 2. (. 2. . Multiply both sides by x +1. ) ( x2+2) to get. x +2. ). (. 2. ). x +x +2x+1=(Ax+B) x +2 +(Cx+D) x +1 . (. 3. ) (. 2. 3. ). 2. x +x +2x+1= Ax +Bx +2Ax+2B + Cx +Dx +Cx+D 3. 2. x +x +2x+1=(A+C)x +(B+D)x +(2A+C)x+(2B+D) . Comparing coefficients gives us the following system of equations: A+C=1 2A+C=2. (1) B+D=1 (3) 2B+D=1. (2) (4). Subtracting equation (1) from equation (3) gives us A=1 , so C=0 . Subtracting equation (2) from equation (4) gives us B=0 , so D=1 . Thus, I=. 3. 2. x +x +2x+1. ( x +1) ( x +2) 2. 2. dx=. x 2. x +1. +. 1 2. dx . For. x +2 6.
(71) Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.4 Integration of Rational Functions by Partial Fractions. . x 2. dx , let u=x +1 so du=2xdx and then . 2. dx , use Formula 10 with a= 2 . So . x. 2. x +1 1. x +2 1 1 2 1 x I= ln x +1 + tan +C . 2 2 2. (. 2. 2. dx=. x +1 1 x +2. (. 1 2. x +( 2. ). 2. dx=. ). 1 1 x tan +C . Thus, 2 2. ). 2. x 2x1. 28.. 1 1 1 1 2 du= ln u +C= ln x +1 +C . For 2 u 2 2. dx=. 2. ( x1 ). ( x2+1). =. A + x1. B 2. +. Cx+D 2. (. 2. 2. ) (. ). 2. 2. x 2x1=A(x1) x +1 +B x +1 +(Cx+D) ( x1 ) .. x +1. (x1). 3. Setting x=1 gives B=1 . Equating the coefficients of x gives A=C . Equating the constant terms gives 1= A1+D , so D=A , and setting x=2 gives 1=5A52A+A or A=1 . We have. . 2. x 2x1. ( x1 ). 2. ( x +1) 2. dx =. 1 x1. . = ln x1 +. 1 2. x1. . dx. 2. x +1. (x1). 1 1 2 1 ln x +1 +tan x+C x1 2. (. ). 29.. . x+4 2. x +2x+5. 3. dx =. 2. x +2x+5. 2. 4. dx+. 3. dx=. 2. 3dx 1 (2x+2)dx + 2 2 2 x +2x+5 (x+1) +4. x +2x+5 2du 1 2 where x+1=2u, = ln x +2x+5 +3 2 and dx=2du 2 4(u +1) 1 3 1 3 2 1 2 1 = ln (x +2x+5)+ tan u+C= ln (x +2x+5)+ tan 2 2 2 2. x 2x +x+1. 30.. x+1. 2. 3. =. 2. x 2x +x+1. (. 2. )(. ). 2. =. Ax+B 2. +. Cx+D 2. 3. 2. x+1 2. (. 2. +C. ). (. 2. x 2x +x+1= ( Ax+B ) x +4 + (Cx+D ) x +1. ). x +5x +4 x +1 x +4 x +1 x +4 . Equating coefficients gives A+C=1 , B+D=2 , 4A+C=1 , 4B+D=1 A=0 , C=1 , B=1 , D=3 . Now. . 3. 2. x 2x +x+1. 31.. 4. 2. x +5x +4 1 3. x 1. =. dx. dx=. 2. x +1. +. 1. (. 2. (x1) x +x+1. ). =. x3 2. x +4. 1. dx=tan x+. 1 3 2 1 ln x +4 tan (x/2)+C . 2 2. (. ). A Bx+C 2 + 2 1=A x +x+1 +(Bx+C)(x1) . x1 x +x+1. (. ). 7.
(72) Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.4 Integration of Rational Functions by Partial Fractions. 1 2 . Equating coefficients of x and then comparing the constant terms, we get 3 1 1 1 2 0= +B , 1= C , so B= , C= 3 3 3 3 1 1 2 1 x 3 dx = 3 dx+ 3 3 dx= 1 ln x1 1 x+2 dx 2 2 x 1 x1 3 3 x +x+1 x +x+1 1 1 x+1/2 1 ( 3/2 ) dx dx = 3 ln x1 3 2 2 3 x +x+1 ( x+1/2 ) +3/4 1 x+ 2 1 1 2 2 1 = 1 ln x1 ln x +x+1 tan +K 3 6 2 3 3 2 1 1 1 1 2 1 ln x1 ln x +x+1 tan (2x+1) +K = 6 3 3 3 Take x=1 to get A=. (. ). (. ). .. /. 32. 1. 1. . x 2. x +4x+13. 0. 1 (2x+4) 2. x +4x+13 dx2 (x+2)dx +9. dx =. 1. 2. 1 = 2. 2. 0. 0. 18. 1. 1 18 2 1 ln y tan u 13 3 2 1 18 2 1 = ln + tan 2 13 6 3 =. 3. 2. dy 3 2 y 2 du 9u +9 13 2/3. (. 2. ). 1 ln u 3. 204. 2. where y=x +4x+13,dy=(2x+4)dx, x+2=3u, and dx=3du 1. =. 2/3. 1 18 2 ln 2 13 3. 1 tan 4. 2 3. 2 3. 33. Let u=x +3x +4 . Then du=3 x +2x dx 5. 2. 2. x +2x 3. 2. x +3x +4 x. 3. 1 dx= 3. 204. . 24. du. u=. 3 ( x +1 ) 1 = =1. 24. =. 1 1 204 1 17 (ln 204ln 24)= ln = ln . 3 3 24 3 2. A Bx+C 2 + 1=A x x+1 +(Bx+C)(x+1) . Equate the 3 3 3 2 x+1 x +1 x +1 x +1 x x+1 terms of degree 2 , 1 and 0 to get 0=A+B , 0= A+B+C , 1=A+C . Solve the three equations to get 34.. 1. =1. (. ). 8.
(73) Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.4 Integration of Rational Functions by Partial Fractions. 1 1 2 , B= , and C= . So 3 3 3 1 1 2 3 x x 3 dx = 1 3 + 3 dx 3 2 x+1 x +1 x x+1 2x1 1 1 1 dx x ln x+1 + 2 6 2 3 = x x+1. A=. . dx 1 2 3 x + 2 4 1 1 1 1 2 1 tan ( 2x1 ) = x ln x+1 + ln x x+1 6 3 3 3. (. 1. 35.. 4. 2. 1. =. x x. =. 2. x (x1)(x+1). A. +. x. B 2. ). C. +. +. x1. x. D. +K. 2. . Multiply by x (x1)(x+1) to get. x+1. 2. 2. 1=Ax(x1)(x+1)+B(x1)(x+1)+Cx (x+1)+Dx (x1) . Setting x=1 gives C=. 1 , taking x=1 gives 2. 1 3 . Equating the coefficients of x gives 0=A+C+D=A . Finally, setting x=0 yields B=1 . Now 2 x1 1 1 dx 1 1/2 1/2 = + dx= + ln +C . 4 2 2 2 x1 x+1 x+1 x x x x. D=. . 4. (. 2. ). 3. (. 3. ). 36. Let u=x +5x +4 du= 4x +10x dx=2 2x +5x dx , so 1. 0. 3. 2x +5x 4. 2. x +5x +4. 37. . 1 dx= 2. x3. ( x +2x+4) 2. 2. 10. . 4. du 1 = ln u u 2. dx=. 10 4. x3 2. ( x+1 ) +3. 2. =. 1 1 5 (ln 10ln 4)= ln . 2 2 2. dx=. u4. ( u +3) 2. 2. du [ with u=x+1 ]. 2. 2 3sec d 1 dv v=u +3 in the first integral; 4 = 4 = 4 2 2 2 2 2 2 u= 3 tan in the second 9sec v u +3 u +3 4 3 2 3 1 1 2 = cos d = ( +sin cos ) +C 2 9 9 ( 2v ) 2 u +3 2 3 3 (x+1) 1 x+1 1 = tan + +C 2 2 9 3 x +2x+4 2 x +2x+4. udu. (. du. ). (. ). (. (. ). ). 9.
(74) Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.4 Integration of Rational Functions by Partial Fractions. 1. =. 2 3 1 tan 9. =. 2 A Bx+C 4 2 2 Dx+E + 2 + x +1=A x +1 +(Bx+C)x x +1 +(Dx+E)x . Setting x=0 2 x 2 x +1 x +1. 2 ( x +2x+4) 2. 4. x +1. 38.. (. 2. x x +1. ). 2. x+1 3. . (. . 2(x+1). +C. 2 3 ( x +2x+4). (. ). ). (. ). 4. gives A=1 , and equating the coefficients of x gives 1=A+B , so B=0 . Now C 2. +. x +1. 4. Dx+E. ( x +1) 2. x +1. =. 2. (. 2. x x +1. . 2. ). 1. =. x. (. x x +1. 2. x. x +1 2. 4. x +1 x +2x +1. 4. D=2 , and E=0 . Hence, . (. 4. 1. 2. ). dx=. 2. ( x +1). 2. 1. . 2x. x. ( x +1) 2. ) 2. =. 2x. ( x +1) 2. 2. dx=ln x +. , so we can take C=0 , 1 2. +C .. x +1. 2. 39. Let u= x+1 . Then x=u 1 , dx=2udu dx 2udu du u1 = 2 =2 2 =ln u+1 x x+1 u 1 u u 1. (. ). +C=ln. x+1 1 x+1 +1. +C .. dx 2udu udu and = 2 =2 2 x x+2 u 2u u u2 A B 1 = + u=A(u+1)+B(u2) . Substituting 1 for u gives 1=3B B= and u2 u+1 3. 40. Let u= x+2 . Then x=u 2 , dx=2udu I= 2. u 2. u u2. substituting 2 for u gives 2=3A A=. 2 . Thus, 3. 2 1 2 2 + du= ( 2ln u2 +ln u+1 ) +C u2 u+1 3 3 2 = 2ln x+2 2 +ln ( x+2 +1 ) +C 3. I =. 2. 41. Let u= x , so u =x and dx=2udu . Thus, 16. 9. 4. . x dx = x4. 3. 4. u 2. u 4. . 2udu=2. 3. u 2. 4. 2. u 4. . du=2. 3. 1+. 4 2. u 4. du. [by long division]. 4. =2+8. 3. du. (u+2)(u2). ( * ) 10.
(75) Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.4 Integration of Rational Functions by Partial Fractions. 1 A B = + by (u+2)(u2) to get 1=A(u2)+B(u+2) . Equating coefficients we (u+2)(u2) u+2 u2 1 1 1 1/4 1/4 get A+B=0 and 2A+2B=1 . Solving gives us B= and A= , so and ( * = + 4 (u+2)(u2) u+2 u2 4 ) is Multiply. 4. 1/4 1/4 + u+2 u2. 2+8. 3. . du = 2+8. ln. = 2+2ln 3. 3. 1 2 ln 5 6. =2+2. 3. =2+2ln. ln. 4. u2 u+2. 3. 2/6 1/5. 2. 5 3. 5 or2+ln 3. 3. 4. = 2+ 2ln u2 2ln u+2 = 2+2. 4. 1 1 ln u+2 + ln u2 4 4. =2+ln. 25 9. 2. 42. Let u= x . Then x=u , dx=3u du 1. 0. 1. 1. dx =. 3. 1+ x. 0. 2. = 3. ln 2. 3u3+. 3 1+u. 3 2 u 3u+3ln (1+u) 2. du=. 1 0. 1 2. 2. 2. 3. 2. x +1 . Then x =u 1 , 2xdx=3u du 3 3 2 3 x dx u 1 u du = 2 3 3 5 3 2 4 3 2 = u u du= u u +C x +1 2 10 4 u 5/3 3 2/3 3 2 2 = x +1 x +1 +C 4 10. 43. Let u=. . 3. 1. 3u 1+u= du 0. (. ). (. (. ). (. ). ). 3. 2. 44. Let u= x . Then x=u , dx=2udu. . 1/3 1. =2 tan u. 3 1/ 3. =2. 3 6. =. x 2. x +x. 3. dx=. . 1/ 3. u 2udu 4. u +u. 2. 3. =2. . 1/ 3. du 2. u +1. 3. 45. If we were to substitute u= x , then the square root would disappear but a cube root would remain. On the other hand, the substitution 11.
(76) Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.4 Integration of Rational Functions by Partial Fractions. u=. 3. x would eliminate the cube root but leave a square root. We can eliminate both roots by means 6. of the substitution u= x . (Note that 6 is the least common multiple of 2 and 3 .) 6. 6. 5. 3. Let u= x . Then x=u , so dx=6u du and x =u , dx. . . =. 3. x x. 5. 6u du 3. u u. = 6. . 12. dx 3. =6. x . Then x=u. . 11. 12u du. 12. 6. x 1 +C. 11. , dx=12u du 8. u du 1 7 6 5 4 3 2 =12 =12 u u +u u +u u +u1+ u+1 u+1. du 4 3 u +u 3 8 12 7 6 12 5 4 3 2 = u u +2u u +3u 4u +6u 12u+12ln u+1 +C 7 5 2 3 2/3 12 7/12 12 5/12 3 4 6 12 = x x +2 x x +3 x 4 x +6 x 12 x +12ln 2 7 5. =. 4. x+ x. x. 2x. e dx 2x. 3. u du=6 du u1. 2. 47. Let u=e . Then x=ln u , dx=. . 5. 2. x =u . Thus,. u (u1) 1 2 u +u+1+ du [by long division] u1 1 3 1 2 3 6 u + u +u+ln u1 +C=2 x +3 x +6 x +6ln 3 2. = 6. 46. Let u=. 2. u. 3. =. x. e +3e +2. . 2. u ( du/u ) 2. u +3u+2. =. ( 12 x +1) +C. du u udu = (u+1)(u+2). 1 2 + u+1 u+2. du. = 2ln u+2 ln u+1 +C=ln (ex+2)2 /(ex+1) +C 48. Let u=sin x . Then du=cos xdx cos xdx du du 1 1 2 = 2 = u(u+1) = u u+1 du sin x+sin x u +u sin x u = ln +C=ln +C 1+sin x u+1. (. 2. ). 49. Let u=ln x x+2 , dv=dx . Then du=. 2x1 2. x x+2. dx , v=x , and (by integration by parts) 12.
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