Question

Integral!

Answer 1

nevermind i got it.

Answer 2

Since you got it.. Then I just type my version..
\[\int \frac{cosx}{1+sin^2x}dx\]\[=\int \frac{1}{1+sin^2x}d(sinx)\]\[=\int \frac{1}{1+u^2}du\]\[=tan^{-1}u+C\]\[=tan^{-1}(sinx)+C\]