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.

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

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

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

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

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