Integral Help
I need to find the integral of (sin x)/ cos^3 x
I let u= cos x, then got du= sin x
(Is this right correct?)
I then rewrote the integral as the integral of du/ u^3 and then rewrote that as the integral of  du(u^3). For this part, I wasn't sure how to finish. I was hoping to get some help. I'd really appreciate any input
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Anonymous

u = cosx
du = sinx dx
du/u^3
1/2cos^2x +c
Or
sec^2 x /2 + cposted by Kuai
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