Question

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

Answer 1

try integration by parts

Answer 2

u = sinx dv = sinx
du = cosx v = cosx

sinxcosx - \[\int\limits_{?}^{?} \cos^2x\]

Answer 3

i don't see how thats going to get me anywhere

Answer 4

sorry is it \[\sin ^{2}x\] or \[\sin x ^{2}\]???

Answer 5

the first.

Answer 6

\[\sin ^2x\]

Answer 7

\[\cos ^{2}x=1-\sin ^{2}x\] \[\int\limits_{?}^{?}\sin ^{2}xdx=-\sin x \cos x +\int\limits_{?}^{?}(1-\sin ^{2}x)dx\]

Answer 8

therefore, the last integral of sin^2(x) can be put to the left hand side.

Answer 9

giving integral (sin^2(x))=(-sin(x)cos(x)+x)/2

Answer 10

would the two integrals divided by each other reduce to -1 though? so it'd be =sinxcosx +1 in the end?

Answer 11

\[\int\limits_{?}^{?}\sin^2x\] \[\div\] \[\int\limits_{}^{} 1-\sin^2x\] = -1?

Answer 12

or is that just improper math? I'm not sure, I dont remember ever dividing integrals.

Answer 13

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?

Answer 14

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

Answer 15

Oh, okay, I get it all now I forgot about separating integrals and what not. This all makes sense.

Answer 16

My AP calc test is on wednesday, so, reviewing all this stuff is really helpful, but a pain in the retriceat the same time.

Answer 17

good luck for wednesday.

Answer 18

thanks.