integral question

[caseclosed]

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)

[TD]

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

[caseclosed]

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

[PhDP]

[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]

[Dave]

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

[caseclosed]

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.

[PhDP]

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 !

[Dave]

Corrected it for you