Posted by H on Monday, October 1, 2012 at 10:16pm.
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 step-by-step 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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