## anonymous 5 years ago How do I find the integral of sin(x)^2

1. anonymous

try integration by parts

2. anonymous

u = sinx dv = sinx du = cosx v = cosx sinxcosx - $\int\limits_{?}^{?} \cos^2x$

3. anonymous

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

4. anonymous

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

5. anonymous

the first.

6. anonymous

$\sin ^2x$

7. anonymous

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

8. anonymous

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

9. anonymous

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

10. anonymous

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

11. anonymous

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

12. anonymous

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

13. anonymous

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?

14. anonymous

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

15. anonymous

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

16. anonymous

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

17. anonymous

good luck for wednesday.

18. anonymous

thanks.