# Integration by parts.

• April 7th 2009, 03:14 PM
gammaman
Integration by parts.
I almost understand all of this question, except in the last step where we have "128". I will show my work up to that step so you can follow.

$\int{x^2sin\frac{x}{4}}dx$

$-4x^2cos\frac{1}{4}x--8\int{xcos\frac{1}{4}}xdx$

$-4x^2cos\frac{1}{4}x+8[{4xsin\frac{1}{4}x-4\int{sin\frac{1}{4}xdx}}]$

$-4x^2cos\frac{1}{4}x+32xsin\frac{1}{4}x+128cos\frac {1}{4}x+c$

I thought that distributing the 8 in step 3 would give us "+32cos...x+c?"
• April 7th 2009, 03:24 PM
e^(i*pi)
Quote:

Originally Posted by gammaman
I almost understand all of this question, except in the last step where we have "128". I will show my work up to that step so you can follow.

$\int{x^2sin\frac{x}{4}}dx$

$-4x^2cos\frac{1}{4}x--8\int{xcos\frac{1}{4}}xdx$

$-4x^2cos\frac{1}{4}x+8[{4xsin\frac{1}{4}x-4\int{sin\frac{1}{4}xdx}}]$

$-4x^2cos\frac{1}{4}x+32xsin\frac{1}{4}x+128cos\frac {1}{4}x+c$

I thought that distributing the 8 in step 3 would give us "+32cos...x+c?"

It comes from dividing the 32 by the 1/4 in sin(x/4) to give 128
• April 7th 2009, 03:28 PM
gammaman
Ok I see it now. The anti-derivative of sin[(1/4)] is -4cos so -4 x -4 = +16
16 x the 8 in front gives us 128.