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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

1 answer

u = cosx

du = -sinx dx

-du/u^-3

1/2cos^2x +c

Or
sec^2 x /2 + c
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