# Math Help - Trig integration

1. ## Trig integration

I'm having a bit of trouble integrating:

$\int(sin(2x) + cos(2x))^2 \ dx$. According to the solution in the back of my book, it's supposed to be $x - (1/4)cos(4x) + C$.

I was thinking of using integration by parts with my $dv = (sin(2x) + cos(2x))$. Anybody have any ideas on how to solve this?

2. Use the identity $(sin(2x)+cos(2x))^{2}=sin(4x)+1$

Then $\int\left[sin(4x)+1\right]dx$

Easier now?.

3. thanks, but how did you simplify it to $\int\left[sin(4x)+1\right]dx$ ?

4. Expand it out and you can see it.

$sin^{2}(2x)+cos^{2}(2x)+2sin(2x)cos(2x)$

If you know your identites you can see that $sin^{2}(2x)+cos^{2}(2x)=1$ and since $sin(2x)=2sinxcosx$ then
$2sin(2x)cos(2x)=sin(4x)$