Prove-that-9-1-ln-1-x-1-x-2-dx-pi-ln-2-8-

Question Number 163536 by HongKing last updated on 07/Jan/22

Prove that :

\underset{9}{\int^{1}} \frac{\ln (1 + x)}{1 + x^{2}} dx = \frac{π \cdot \ln (2)}{8}

Answered by Ar Brandon last updated on 07/Jan/22

I = \int_{0}^{1} \frac{\ln (1 + x)}{1 + x^{2}} d x, x = \tan ϑ

= \int_{0}^{\frac{π}{4}} \ln (1 + \tan ϑ) d ϑ

= \int_{0}^{\frac{π}{4}} \ln (\cos ϑ + \sin ϑ) d ϑ - \int_{0}^{\frac{π}{4}} \ln (\cos ϑ) d ϑ

= \int_{0}^{\frac{π}{4}} \ln (\sqrt{2} \cos (x - \frac{π}{4})) d x - \int_{0}^{\frac{π}{4}} \ln (\cos ϑ) d ϑ

= \frac{π \ln 2}{8} + \int_{0}^{\frac{π}{4}} \ln (\cos (x - \frac{π}{4})) - \int_{0}^{\frac{π}{4}} \ln (\cos ϑ) d ϑ

= \frac{π \ln 2}{8} + \int_{0}^{\frac{π}{4}} \ln (\cos ϑ) d ϑ - \int_{0}^{\frac{π}{4}} \ln (\cos ϑ) d ϑ = \frac{π \ln 2}{8}

Commented by HongKing last updated on 07/Jan/22

perfect my dear Sir thank you so much

Commented by peter frank last updated on 11/Jan/22

great

Answered by smallEinstein last updated on 07/Jan/22

Commented by GalaxyBills last updated on 07/Jan/22

M y B o s s t h a t