Question

Integrate cosx/(1+sin^2x)?

Answer 1

We have:

I=∫cosx/1+sin²xdx

Let sinx=tanθ. This may look like a wild substitution, but the goal is to get the denominator of the fraction to 1+tan²θ=sec²θ. Note that differentiating sinx=tanθ gives cosxdx=sec²θdθ. It's also helpful that cosxdx is already present in the integrand. Substituting these in:

I=∫sec²θ/1+tan²θdθ=∫dθ = θ+C

Since sinx=tanθ, we see that θ=arctan(sinx):

I=arctan(sinx)+C