How do you find the integral of int (sin x)/(cos^2x + 1)dxsinxcos2x+1dx?

1 Answer
Oct 20, 2015

I = -arctan (cosx) + CI=arctan(cosx)+C

Explanation:

I = int sinx/(cos^2x+1)dx = int (sinxdx)/(cos^2x+1) = -int (d(cosx))/(cos^2x+1)I=sinxcos2x+1dx=sinxdxcos2x+1=d(cosx)cos2x+1

I = -arctan (cosx) + CI=arctan(cosx)+C