integrate sin(4x)^2

 

I know

sin(x)^2=1-cos(x)^2

cos(x)^2=sin(x)^2+cos(2x)

 

ok I am lost, problem is sin(4x) so but the identities are in sin(x)

The problem isn't the 4x, but the square! Since you can easily integrate sin(x), it's no problem to integrate sin(ax) with a an arbitrary constant either!

 

To get rid of the square, use another trig identity:

 

[math]\cos \left( {2x} \right) = \cos ^2 x - \sin ^2 x = 1 - 2\sin ^2 x \Leftrightarrow \sin ^2 x = \frac{{1 - \cos \left( {2x} \right)}}{2}[/math]

 

Now you lost the square, went to a cosine and doubled the angle - but that shouldn't be a problem

umm, sin(x)^2= (1-cos(x))/2

but what does sin(4x)^2=?

I am very confused on what to do with stuff inside the trig function since I can't pull it out in front

[math]\sin^2(x) = \frac{1-\cos(2x)}{2}[/math]

 

You just have to replace the "x" in the identity with your "4x";

 

[math]\sin^2(4x) = \frac{1-\cos(2(4x))}{2}=\frac{1-\cos(8x)}{2}[/math]

 

So your integral is now;

 

[math]\int \sin^2(4x)dx = \frac{1}{2}\int dx - \frac{1}{2}\int \cos(8x)dx[/math]

Phil: I think you may have meant [imath]\sin^2 4x[/imath] instead of [imath]\sin^2 x[/imath] in your last integral.

 

As for evaluating something like [math]\int\cos(ax)\, dx[/math], note the following:

 

[math]\frac{d}{dx} \sin(ax) = a\cos(ax)[/math]

 

Now, integrate both sides to get: [math]\sin(ax) = a\int\cos(ax)\, dx[/math]

 

You should be able to see where to go from here

You just have to replace the "x" in the identity with your "4x";

 

thanks, that is the part I was really confused on for a really really long time.

Phil: I think you may have meant \sin^2 4x instead of \sin^2 x in your last integral.

 

Exactly, but I can't correct it now, we're doomed !

Corrected it for you