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Topic: Calculus Problem (Integration by Parts)
Replies: 1   Last Post: Jul 17, 1996 1:20 PM

 Messages: [ Previous | Next ]
 Stephen Pride Posts: 2 Registered: 12/6/04
Calculus Problem (Integration by Parts)
Posted: Jul 12, 1996 4:35 PM

I am trying to work a problem in Calculus and keep arriving at an answer
that is different from the book. Can someone please "enlighten" me?
Note: "S" stands for the integral sign:

S (e^2x) (sin 3x)

Here is what I am doing:

u = sin 3x dv = e^2x
du = 3 cos 3x v = (e^2x)/2

S (e^2x) (sin 3x) dx = (sin 3x) (e^2x)/2 - S (e^2x)/2 (3 cos 3x)
dx

" = (e^2x)/2 (sin 3x) - (3/2) S (e^2x) (cos 3x) dx

u = cos 3x dv = e ^2x
du = 3 sin x v = (e^2x)/2

" = (e^2x)/2 (sin 3x) - (3/2) [ (cos 3x) (e^2x)/2 - S (e^2x)/2
(3 sin 3x) dx ]

" = (e^2x)/2 (sin 3x) - (3 e^2x)/4 (cos 3x) + (9/4) S (e^2x)
(sin 3x) dx

(subtract: (9/4) S (e^2x) (sin 3x) dx, from both sides)

(-5/4) S (e^2x) (sin 3x) = (e^2x)/2 (sin 3x) - (3 e^2x)/4 (cos
3x)

(multiply both sides by: (-4/5) )

S (e^2x) (sin 3x) dx = (-2/5) (e^2x) (sin 3x) + (3/5) (e^2x)
(cos 3x) + C

But the book has the following for the answer:

S (e^2x) (sin 3x) dx = (e^2x) (2 sin 3x - 3 cos 3x)/13 + C

Thanks for any help,
steve

Date Subject Author
7/12/96 Stephen Pride
7/17/96 Damon C Capehart