1. Integration with Trigonometric Functions

Here's what I have so far, but I'm coming up with answers other than the book value. Mind pointing out my error?

$\int sin2xcos2xdx$

Letting $u=sin2x, du=cos2xdx$

$\int udu$
$\frac{u^2}{2}+C$
$\frac{sin^{2}2x}{2}+C$

2. Originally Posted by xfriendsonfirex
Here's what I have so far, but I'm coming up with answers other than the book value. Mind pointing out my error?

$\int sin2xcos2xdx$

Letting $u=sin2x, du=cos2xdx$

$\int udu$
$\frac{u^2}{2}+C$
$\frac{sin^{2}2x}{2}+C$
If $u=sin(2x)$, then $du={\color{red}2}cos(2x)dx$.

3. Could you walk me through why that is true?

4. Surely, you learned the chain rule. Right?

5. Originally Posted by xfriendsonfirex
Could you walk me through why that is true?

chain rule.

$sin(2x)$ is really the composition of $f(x)=sin(x)$ and $g(x)=2x$, so that $f(g(x)) = sin(2x)$. this then means that $f(g(x))^{'} = f^{'}(g(x))g^{'}(x)$ by the chain rule. So $f(g(x))^{'} = f^{'}(g(x))g^{'}(x) = cos(2x)dx*2$

6. It has been a while. Haha.

I just got the gist of it from Google, but a more problem specific view would surely help.

7. I think I got it down. Thanks guys.