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25. First, we evaluate E e 2 t cos t dt . Let u 5 e 2 t dv 5 cos t dt du 52 e 2 t dt v 5 sin t E e 2 t cos t dt 5 e 2 t sin t 1 E sin t e 2 t dt Let u 5 e 2 t dv 5 sin t dt du e 2 t dt v cos t E e 2 t cos t dt 5 e 2 t sin t 2 e 2 t cos t 2 E e 2 t cos t dt 2 E e 2 t cos t dt 5 e 2 t (sin t 2 cos t ) 1 C E e 2 t cos t dt 5 } 1 2 } e 2 t (sin t 2 cos t ) 1 C Now we find the average value of y 5 2 e 2 t cos t for 0 # t # 2 p . Average value 5 } 2 1 p } E 2 p 0 2 e 2 t cos t dt 5 } p 1 } E 2 p 0 e 2 t cos t dt 5 } 2 1 p } e 2 t (sin t 2 cos t ) 4 5 } 2 1 p } [ e 2 2 p ( 2 1) 2 e 0 ( 2 1)] 5 } 1 2 2 p e 2 2 p } < 0.159 26. (a) Let u 5 xd v 5 e x dx du 5 dx v 5 e x E xe x dx 5 xe x 2 E e x dx 5 xe x 2 e x 1 C 5 ( x 2 1) e x 1 C (b) Using the result from part (a): Let u 5 x 2 dv 5 e x dx du 5 2 x dx v 5 e x E x 2 e x dx 5 x 2 e x 2 E 2 xe x dx 5 x 2 e x 2 2( x 2 1) e x 1 C 5 ( x 2 2 2 x 1 2) e x 1 C (c) Using the result from part (b): Let u 5 x 3 dv 5 e x dx du 5 3 x 2 dx v 5 e x E x 3 e x dx 5 x 3 e x 2 E 3 x 2 e x dx 5 x 3 e x 2 3( x 2 2 2 x 1 2) e x 1 C 5 ( x 3 2 3 x 2 1 6 x 2 6) e x 1 C (d) 3 x n 2 } d d x } x n 1 } d d x 2 2 } x n 2 1 ( 2 1) n } d d x n n } x n 4 e x 1 C or [ x n 2 nx n 2 1 1 n ( n 2 1) x n 2 2 2 1 ( 2 1) n 2 1 ( n 2 1)! x 1 ( 2 1) n ( n !)] e x 1 C (e) Use mathematical induction or argue based on tabular integration. Alternately, show that the derivative of the answer to part (d) is x n e x : } d d x } 31 x n 2 nx n 2 1 1 n ( n 2 1) x n 2 2 2 1 ( 2 1) n 2 1 ( n !) x 1 ( 2 1) n n ! 2 e x 1 C 4 5 [ x n 2 nx n 2 1 1 n ( n 2 1) x n 2 2 2 1 ( 2 1) n 2 1 ( n !) x 1 ( 2 1) n n !] e x 1 e x } d d x } [ x n 2 nx n 2 1 1 n ( n 2 1) x n 2 2 2 1 ( 2 1) n 2 1 ( n !) x 1 ( 2 1) n n !] 5 [ x n 2 nx n 2 1 1 n ( n 2 1) x n 2 2 2 1 ( 2 1) n 2 1 ( n !) x 1 ( 2 1) n n !] e x 1 [ nx n 2 1 2 n ( n 2 1) x n 2 2 1 n ( n 2 1)( n 2 2) x n 2 3 2 1 ( 2 1) n 2 1 n !] e x 5 x n e x 27. Let w 5 ˇ x w . Then dw 5 } 2 d ˇ x x w } ,so dx 5 2 ˇ x w dw 5 2 w dw . E sin ˇ x w dx 5 E (sin w )(2 w dw ) 5 2 E w sin w dw Let u 5 wd v 5 sin w dw du 5 dw v cos w E w sin w dw w cos w 1 E cos w dw w cos w 1 sin w 1 C E sin ˇ x w dx 5 2 E w sin w dw 2 w cos w 1 2 sin w 1 C 2 ˇ x w cos ˇ x w 1 2 sin ˇ x w 1 C 2 p 0 256 Section 6.3

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28. Let w 5 ˇ 3 w x w 1 w 9 w . Then dw 5 } 2 ˇ 3 w 1 x w 1 w 9 w } (3) dx , so dx 5 } 2 3 } ˇ 3 w x w 1 w 9 w dw 5 } 2 3 } w dw . E e ˇ 3 w x 1 w 9 w dx 5 E ( e w ) 1 } 2 3 } w dw 2 5 } 2 3 } E we w dw Let u 5 wd v 5 e w dw du 5 dw v 5 e w E w dw 5 w 2 E e w dw 5 w 2 e w 5 ( w 2 1) e w E e ˇ 3 w x 1 w 9 w dx 5 } 2 3 } E w dw 5 } 2 3 } ( w 2 1) e w 5 } 2 3 } ( ˇ 3 w x w 1 w 9 w 2 1) e ˇ 3 w x 1 w 9 w 1 C 29. Let w 5 x 2 . Then dw 5 2 x dx .
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