Question

Please help? Integrals?

Answer 1

How would I find the integral of \[\int\limits_{}^{}\frac{ \cos x }{ \sin^2x }\]

Answer 2

What's stopping you from doing a u sub :P

Answer 3

I've tried substituting sin^2x in for u and I'm confused because then I would du= sin 2x dx and then I'm confused about how that wold work in a fraction

Answer 4

So would I substitute u=cos x?

Answer 5

No, here's the trick, notice you have a cosx in the numerator, and a sin^2x, so what you want to do is let u = sinx

Answer 6

that way cosx will get eliminated

Answer 7

\[\sin^2x+\cos^2x=1\]
therefore:
\[\sin^2x=1-\cos^2x\]