Question Number 149718 by peter frank last updated on 06/Aug/21

Answered by Ar Brandon last updated on 06/Aug/21

I=∫cos2θln(((cosθ+sinθ)/(cosθ−sinθ)))dθ =∫cos2θln(((1+tanθ)/(1−tanθ)))dθ=∫cos2θln(tan((π/4)+θ))dθ { ((u(x)=ln(tan((π/4)+θ)))),((v′(x)=cos2θ)) :}⇒ { ((u′(x)=(2/(sin((π/2)+2θ))))),((v(x)=(1/2)sin2θ)) :} I=(1/2)sin2θ∙ln(tan((π/4)+θ))−∫((sin2θ)/(cos2θ))dθ =(1/2)sin2θ∙ln(tan((π/4)+θ))+(1/2)ln(cos2θ)+C