# integral question

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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)

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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:

$\cos \left( {2x} \right) = \cos ^2 x - \sin ^2 x = 1 - 2\sin ^2 x \Leftrightarrow \sin ^2 x = \frac{{1 - \cos \left( {2x} \right)}}{2}$

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

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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

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$\sin^2(x) = \frac{1-\cos(2x)}{2}$

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

$\sin^2(4x) = \frac{1-\cos(2(4x))}{2}=\frac{1-\cos(8x)}{2}$

$\int \sin^2(4x)dx = \frac{1}{2}\int dx - \frac{1}{2}\int \cos(8x)dx$

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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 $\int\cos(ax)\, dx$, note the following:

$\frac{d}{dx} \sin(ax) = a\cos(ax)$

Now, integrate both sides to get: $\sin(ax) = a\int\cos(ax)\, dx$

You should be able to see where to go from here

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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.

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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 !

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Corrected it for you

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