Let $x=-t$, so $dx=-dt$

$I = \int\limits_\pi ^{ – \pi } { – \frac{{{{\sin }^2}t}}{{1 + {e^{ – t}}}}dt} = \int\limits_{ – \pi }^\pi {\frac{{{e^t}{{\sin }^2}t}}{{1 + {e^t}}}dt} = \int\limits_{ – \pi }^\pi {\frac{{{e^x}{{\sin }^2}x}}{{1 + {e^x}}}dx} $

$I + I = \int\limits_{ – \pi }^\pi {\frac{{{{\sin }^2}x}}{{1 + {e^x}}}dx} + \int\limits_{ – \pi }^\pi {\frac{{{e^x}{{\sin }^2}x}}{{1 + {e^x}}}dx} $

$ \Rightarrow 2I = \int\limits_{ – \pi }^\pi {\frac{{(1 + {e^x}){{\sin }^2}x}}{{1 + {e^x}}}dx} = \int\limits_{ – \pi }^\pi {{{\sin }^2}xdx} $

$ \Rightarrow 2I = \int\limits_0^\pi {2{{\sin }^2}xdx} = \int\limits_0^\pi {(1 – \cos 2x)dx} $

$ \Rightarrow 2I = \left. {\left( {x – \frac{{\sin 2x}}{2}} \right)} \right|_0^\pi = \pi $

$ \Rightarrow I = \frac{\pi }{2}$

Let, $x = \tan \theta $

$ \Rightarrow dx = {\sec ^2}\theta d\theta $

$I = \int\limits_0^{\pi /4} {\frac{{\ln (1 + \tan \theta )}}{{1 + {{\tan }^2}\theta }}{{\sec }^2}\theta d\theta } $

$ = \int\limits_0^{\pi /4} {\ln \left( {\frac{{\sin \theta + \cos \theta }}{{\cos \theta }}} \right)d\theta } $

$ = \int\limits_0^{\pi /4} {\ln (\sin \theta + \cos \theta )d\theta – \int\limits_0^{\pi /4} {\ln \cos \theta d\theta } } $

$ = \int\limits_0^{\pi /4} {\ln \left\{ {\sqrt 2 \left( {\cos \theta .\frac{1}{{\sqrt 2 }} + \sin \theta .\frac{1}{{\sqrt 2 }}} \right)} \right\}d\theta – \int\limits_0^{\pi /4} {\ln \cos \theta d\theta } } $

$ = \int\limits_0^{\pi /4} {\ln \left\{ {\sqrt 2 \left( {\cos \theta .\cos \frac{\pi }{4} + \sin \theta .\sin \frac{\pi }{4}} \right)} \right\}d\theta – \int\limits_0^{\pi /4} {\ln \cos \theta d\theta } } $

$ = \int\limits_0^{\pi /4} {\ln \left\{ {\sqrt 2 \cos \left( {\theta – \frac{\pi }{4}} \right)} \right\}d\theta – \int\limits_0^{\pi /4} {\ln \cos \theta d\theta } } $

$ = \int\limits_0^{\pi /4} {\ln \sqrt 2 d\theta + } \int\limits_0^{\pi /4} {\ln \cos \left( {\theta – \frac{\pi }{4}} \right)d\theta – \int\limits_0^{\pi /4} {\ln \cos \theta d\theta } } $

Let, $\theta – \frac{\pi }{4} = – \phi $ for the second integral.

$I = \frac{\pi }{4}\ln \sqrt 2 – \int\limits_{\pi /4}^0 {\ln \cos ( – \phi )d\phi – \int\limits_0^{\pi /4} {\ln \cos \theta d\theta } } $

$ = \frac{\pi }{4}\ln \sqrt 2 + \int\limits_0^{\pi /4} {\ln \cos \phi d\phi – \int\limits_0^{\pi /4} {\ln \cos \theta d\theta } } $

$ = \frac{\pi }{4}\ln \sqrt 2 = \frac{\pi }{8}\ln 2$

Let, ${e^{ – x}} = t$

$ \Rightarrow  – {e^{ – x}}dx = dt$

$ \Rightarrow  – tdx = dt$

$I = \int\limits_1^0 {\frac{{{t^n} – t}}{{ – \ln t}}\frac{{dt}}{{ – t}}}  =  – \int\limits_0^1 {\frac{{{t^{n – 1}} – 1}}{{\ln t}}dt} $

But, $I(\lambda ) = \int\limits_0^1 {\frac{{{x^\lambda } – 1}}{{\ln x}}dx = \ln (\lambda  + 1)} $ (Proof can be seen in this problem)

So, $I =  – \ln (n – 1 + 1) =  – \ln n$

$I = \int\limits_0^\pi {\frac{{2(\pi – x)\sin (\pi – x)}}{{3 + \cos 2(\pi – x)}}dx} $

$ = \int\limits_0^\pi {\frac{{2(\pi – x)\sin x}}{{3 + \cos 2x}}dx} $

$2I = \int\limits_0^\pi {\frac{{2x\sin x}}{{3 + \cos 2x}}dx} + \int\limits_0^\pi {\frac{{2(\pi – x)\sin x}}{{3 + \cos 2x}}dx} = \int\limits_0^\pi {\frac{{2\pi \sin x}}{{3 + \cos 2x}}dx} $ Continue reading $I = \int\limits_0^\pi {\frac{{2x\sin x}}{{3 + \cos 2x}}dx} = ?$ →