How do I find the integral of sin(x)^2

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How do I find the integral of sin(x)^2

Mathematics
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try integration by parts
u = sinx dv = sinx du = cosx v = cosx sinxcosx - \[\int\limits_{?}^{?} \cos^2x\]
i don't see how thats going to get me anywhere

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sorry is it \[\sin ^{2}x\] or \[\sin x ^{2}\]???
the first.
\[\sin ^2x\]
\[\cos ^{2}x=1-\sin ^{2}x\] \[\int\limits_{?}^{?}\sin ^{2}xdx=-\sin x \cos x +\int\limits_{?}^{?}(1-\sin ^{2}x)dx\]
therefore, the last integral of sin^2(x) can be put to the left hand side.
giving integral (sin^2(x))=(-sin(x)cos(x)+x)/2
would the two integrals divided by each other reduce to -1 though? so it'd be =sinxcosx +1 in the end?
\[\int\limits_{?}^{?}\sin^2x\] \[\div\] \[\int\limits_{}^{} 1-\sin^2x\] = -1?
or is that just improper math? I'm not sure, I dont remember ever dividing integrals.
oh wait, I wouldnt be dividng them though, I'd subtract, though I still think I'm missing something where does the 2 in -sinxcosx/2 come from?
no, because you need to add integral (sin^2(x)) to both sides, not dividing them, giving 2 times integral (sin^2(x)) on the LHS and cancelling it on the RHS
Oh, okay, I get it all now I forgot about separating integrals and what not. This all makes sense.
My AP calc test is on wednesday, so, reviewing all this stuff is really helpful, but a pain in the retriceat the same time.
good luck for wednesday.
thanks.

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