Hi experts,

How does one solve

$\displaystyle -\int\frac{sinx}{cos^2x}dx$

I have attempted to use the identity cos^2x=1/2(1+cos2x) to give

$\displaystyle -\int\frac{sinx}{\frac{1}{2}(1+cos2x)}$

However I am not sure how to proceed any further because of the double angle....

In another attempt, i tried tackling it using the method of substitution such as below

let u=cosx then du = -sinxdx

then

$\displaystyle \int\frac{du}{u^2} = -\frac{1}{u} + c$

But this answer doesnt look right either....

Any tips or suggestions on what I am doing wrong will be appreciated. Thanks

Bugatti