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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:

 

[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 :)

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[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]

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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 [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 :)

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