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Question Number 211880 by Spillover last updated on 23/Sep/24

Answered by Frix last updated on 23/Sep/24

$$\underset{0}{\stackrel{\frac{\pi }{2}}{\int }}\frac{\tan x}{{\pi }^2+{4\ln }^2\tan x}dx\stackrel{[t=2\ln \tan x]}{=}$$$$=\frac{1}{2}\underset{-\mathrm{\infty }}{\stackrel{\mathrm{\infty }}{\int }}\frac{{\mathrm{e}}^t}{(t^2+{\pi }^2)({\mathrm{e}}^t+1)}dt\stackrel{[t=-u]}{=}$$$$=-\frac{1}{2}\underset{\mathrm{\infty }}{\stackrel{-\mathrm{\infty }}{\int }}\frac{du}{(u^2+{\pi }^2)({\mathrm{e}}^u+1)}=$$$$=\frac{1}{2}\underset{-\mathrm{\infty }}{\stackrel{\mathrm{\infty }}{\int }}\frac{du}{(u^2+{\pi }^2)({\mathrm{e}}^u+1)}=\frac{1}{2}\underset{-\mathrm{\infty }}{\stackrel{\mathrm{\infty }}{\int }}\frac{dt}{(t^2+{\pi }^2)({\mathrm{e}}^t+1)}$$$$\Rightarrow$$$$2\underset{0}{\stackrel{\frac{\pi }{2}}{\int }}\frac{\tan x}{{\pi }^2+{4\ln }^2\tan x}dx=$$$$\frac{1}{2}\underset{-\mathrm{\infty }}{\stackrel{\mathrm{\infty }}{\int }}\frac{{\mathrm{e}}^t}{(t^2+{\pi }^2)({\mathrm{e}}^t+1)}dt+\frac{1}{2}\underset{-\mathrm{\infty }}{\stackrel{\mathrm{\infty }}{\int }}\frac{dt}{(t^2+{\pi }^2)({\mathrm{e}}^t+1)}=$$$$=\frac{1}{2}\underset{-\mathrm{\infty }}{\stackrel{\mathrm{\infty }}{\int }}\frac{dt}{t^2+{\pi }^2}=\frac{1}{2\pi }{\left[{\tan }^{-1}\frac{t}{\pi }\right]}_{-\mathrm{\infty }}^{+\mathrm{\infty }}=\frac{1}{2}$$$$\Rightarrow$$$$\underset{0}{\stackrel{\frac{\pi }{2}}{\int }}\frac{\tan x}{{\pi }^2+{4\ln }^2\tan x}dx=\frac{1}{4}$$

Commented by Spillover last updated on 23/Sep/24

$$correct.thanks.niceapproach$$

Answered by Spillover last updated on 23/Sep/24

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