Prove-that-0-pi-xtan-x-sec-x-tan-x-dx-pi-2-2-pi-

Question Number 152276 by peter frank last updated on 27/Aug/21

Prove that

\int_{0}^{π} \frac{xtan x}{\sec x + \tan x} dx = \frac{π^{2}}{2} - π

Answered by Olaf_Thorendsen last updated on 27/Aug/21

F (x) = \int_{0}^{π} \frac{x \tan x}{\sec x + \tan x} d x

F (x) = \int_{0}^{π} x \tan x (\sec x - \tan x) d x

F (x) = \int_{0}^{π} x (\frac{\sin x}{\cos^{2} x} - (1 + \tan^{2} x) + 1) d x

F (x) = {[x (\sec x - \tan x + x)]}_{0}^{π}

- \int_{0}^{π} (\sec x - \tan x + x) d x

F (x) = π (π - 1)

- {[\ln ∣ \sec x + \tan x ∣ + \ln ∣ \cos x ∣ + \frac{x^{2}}{2}]}_{0}^{π}

F (x) = π (π - 1) - \frac{π^{2}}{2} = \frac{π^{2}}{2} - π