Trig integration

**lllll** · February 16th 2008, 04:35 PM

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?

**galactus** · February 16th 2008, 04:41 PM

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

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

Easier now?.

**lllll** · February 16th 2008, 04:47 PM

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

**galactus** · February 16th 2008, 05:10 PM

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
$2sin(2x)cos(2x)=sin(4x)$

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