Asked by Dave
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???
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???
Answers
Answered by
Steve
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
(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