Question

Evaluate the integral of (sin 2x)/(1+cos^2 x)


1. u=cosx and du=-sinx *dx

2. evaluate the integral of -1/(1+u^2)*du

3. result is -ln(1+u^2)+C


Where did the sin2x dissapear too???

Answer 1

recall that sin2x = 2 sinx cosx

(sin2x)/(1+cos^2x) dx
= 2cosx/(1+cos^2x) sinx dx
= -2u/(1+u^2) du

let v = 1+u^2, so dv = 2u du

= -dv/v