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\(\displaystyle \int sin^{4}x dx\)

so I know:

\(\displaystyle \frac{1}{2} \int 1 - 2cos2x + \frac{1}{2}(1 + cos4x)dx\)

So here I first brought out the 1/2 because it's a constant and it's nasty.

so now I have

\(\displaystyle \frac{1}{4} \int 1 - 2cos2x + 1 + cos4x dx\)

so...Just as I brought 1/2 out can I now precede to take the 2 constant out that is in front of 2cos2x? thus turning it into:

\(\displaystyle \frac{1}{2} \int 1 - cos2x + 1 + cos4x dx\) ?

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Ah wait, I think I forgot to square the denominator of (1 - cos2x/2)^2 because it was sin^4