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Question Number 215496 by MATHEMATICSAM last updated on 08/Jan/25

$${\int }_0^{\pi }\frac{x\tan x}{\sec x+\tan x}dx$$$$\mathrm{Solve}\mathrm{the}\mathrm{integral}.$$

Commented by mr W last updated on 09/Jan/25

$$=\frac{{\pi }^2}{2}-\pi$$

Answered by Frix last updated on 10/Jan/25

$$\int \frac{x\tan x}{\sec x+\tan x}dx=\int \frac{x\sin x}{1+\sin x}dx\stackrel{\left[\mathrm{by}\mathrm{parts}\right]}{=}$$$$=\frac{x(1+x\cos x-\sin x)}{\cos x}-\int (x+\frac{1}{\cos x}-\tan x)dx$$$$\mathrm{Should}\mathrm{be}\mathrm{easy}\mathrm{from}\mathrm{here}...$$

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