Question

The integral of (sin^3x/cosx)dx
I got it down to look like the integral of (sinx/cosx)-cosxsinxdx. Not sure what to do from there...not sure if I'm on the correct path. Thanks!

Answer 1

is it : \[\frac{sin^3(x)}{cos(x)}\]?

Answer 2

how about writing it as
\[\int\limits_{}^{}\frac{sinx}{cosx}*(1-\cos^2x)dx\]
let u=cosx
du=-sinx dx

Answer 3

sin^2 = (1-cos(2x))/2 as well

Answer 4

\[\int\limits_{}^{}\frac{-du}{u}(1-u^2) =-\int\limits_{}^{}(\frac{1}{u}-\frac{u^2}{u}) du\]

Answer 5

\[\frac{sin}{cos}*[\frac{1}{2}-\frac{cos(2x)}{2}]\]

Answer 6

\[-\ln|u|-\frac{u^2}{2}+C=-\ln|cosx|-\frac{(cosx)^2}{2}+C\]

Answer 7

there should be a plus in front of cos^2x/2

Answer 8

and it front of u^2/2

Answer 9

my way might be more of a hassle; cant tell :)

Answer 10

Okay, I will try and work it out both ways :) Thanks you both!