Calc
posted by H .
Evaluate the integral using any method:
(Integral)sec^3x/tanx dx
I started it out and got secx(1tan^2x)/tanx. I know I just have to continue simplifying and finding the integral, but I'm stuck on the next couple of steps.
Also, I have another question witht he same directions:(integral)sin^3/cosx dx. What would I do here?
Thank you!

I did:
sec^3x/tanx
= (1/cos^3 x)(cosx/sinx)
= 1/((sinx)(cos^2 x)
so I went to my favourite integrator page and got
http://integrals.wolfram.com/index.jsp?expr=1%2F%28%28sin%28x%29cos%5E2%28x%29%29&random=false
You should realize that when Wolfram says log(...)
they really mean ln(...)
for your 2nd
http://integrals.wolfram.com/index.jsp?expr=sin%5E3%28x%29%2Fcos%28x%29&random=false 
thanks for the first part. i wanted something that was more stepbystep for the second one.

sin^3/cosx
= (sinx/cos)(sin^2x)
= tanx sin^2 x
then
http://integrals.wolfram.com/index.jsp?expr=%28tan%28x%29%28sin%28x%29%5E2%29&random=false
same as above
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