$I = \int_0^{\pi /2} {\sin 2\,\left( {\frac{\pi }{2} - x} \right)\log \tan \left( {\frac{\pi }{2} - x} \right)\,\,dx} $,
$[\because \int_{0}^{a}{f\,(x)\,dx=\int_{0}^{a}{f\,(a-x)\,dx]}}$
$ = \int_0^{\pi /2} {\,\,\,\,\,\sin 2x\log \cot x\,\,dx} $
$ = - \int_0^{\pi /2} {\,\,\,\,\,\sin 2x\log \tan x\,\,dx} $
$\therefore I = - I\,\,==> 2I = 0$ $ \Rightarrow I = 0.$
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$\int \frac{\left(x^{2}-1\right)+\tan ^{-1}\left(\frac{ x ^{2}+1}{ x }\right)}{\left( x ^{4}+3 x ^{2}+1\right) \tan ^{-1}\left(\frac{ x ^{2}+1}{ x }\right)} dx$ $=\alpha \log _{ e }\left(\tan ^{-1}\left(\frac{ x ^{2}+1}{ x }\right)\right)+\beta \tan ^{-1}\left(\frac{\gamma\left( x ^{2}-1\right)}{ x }\right)$ $+\delta \tan ^{-1}\left(\frac{ x ^{2}+1}{ x }\right)+ C$ है, जहाँ $C$ एक स्वेच्छ अचर है, तो $10(\alpha+\beta \gamma+\delta)$ का मान बराबर है .......... |