Question

How do you find the integral of `int (sin x)/(cos^2x + 1)dx`?

Answer 1

`I = -arctan (cosx) + C`

Explanation:

`I = int sinx/(cos^2x+1)dx = int (sinxdx)/(cos^2x+1) = -int (d(cosx))/(cos^2x+1)`

`I = -arctan (cosx) + C`