Question

Integral of sinx/(cosx+(cosx)^2) dx, how do I approach this?

Answer 1

let u = cosx

Answer 2

\[I=\int\limits \frac{ \sin x~dx }{ \cos x+\cos ^2x }\]
Put cos x=t
- sin x dx=dt
\[=-\int\limits \frac{ dt }{ t+t^2 }=- \int\limits \frac{ dt }{ t \left( 1+t \right) }=-\left[ \int\limits \left\{ \frac{ 1 }{ t \left( 1+0 \right) }+\frac{ 1 }{ -1\left( 1+t \right) } \right\}\right]dt\]
\[=-\int\limits \frac{ dt }{ t }+\int\limits \frac{ dt }{ 1+t }=?\]
i think now you can complete.

Answer 3

Thanks guys, I tried to set u to cosx at first, but wasn't sure how to handle the u+u^2 in the denominator

Answer 4

yw