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

$$\mathrm{Prove}\mathrm{that}$$$${\int }_0^{\pi }\frac{\mathrm{xtan}\mathrm{x}}{\sec \mathrm{x}+\tan \mathrm{x}}\mathrm{dx}=\frac{{\pi }^2}{2}-\pi$$

Answered by Olaf_Thorendsen last updated on 27/Aug/21

$$\mathrm{F}\left(x\right)={\int }_0^{\pi }\frac{x\tan x}{\sec x+\tan x}dx$$$$\mathrm{F}\left(x\right)={\int }_0^{\pi }x\tan x(\sec x-\tan x)dx$$$$\mathrm{F}\left(x\right)={\int }_0^{\pi }x(\frac{\sin x}{{\cos }^{2}x}-(1+{\tan }^{2}x)+1)dx$$$$\mathrm{F}\left(x\right)={\left[x(\sec x-\tan x+x)\right]}_0^{\pi }$$$$-{\int }_0^{\pi }(\sec x-\tan x+x)dx$$$$\mathrm{F}\left(x\right)=\pi (\pi -1)$$$$-{[\ln \mid \sec x+\tan x\mid +\ln \mid \cos x\mid +\frac{x^2}{2}]}_0^{\pi }$$$$\mathrm{F}\left(x\right)=\pi (\pi -1)-\frac{{\pi }^2}{2}=\frac{{\pi }^2}{2}-\pi$$

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