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

$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} $

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

$ = \pi \int\limits_0^\pi  {\frac{{\sin x}}{{3 + 2{{\cos }^2}x - 1}}dx} $

$ = \frac{\pi }{2}\int\limits_0^\pi  {\frac{{\sin x}}{{1 + {{\cos }^2}x}}dx} $

Let, $\cos x = t$

$ \Rightarrow  - \sin xdx = dt$

$I = \frac{\pi }{2}\int\limits_1^{ - 1} {\frac{{ - dt}}{{1 + {t^2}}}} $

$ = \frac{\pi }{2}\int\limits_{ - 1}^1 {\frac{{dt}}{{1 + {t^2}}}} $

$ = \frac{{\pi  \times 2}}{2}\int\limits_0^1 {\frac{{dt}}{{1 + {t^2}}}} $

$ = \pi \left. {{{\tan }^{ - 1}}t} \right|_0^1$

$ = \frac{{{\pi ^2}}}{4}$

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${\log _{\sqrt 5 }}\left[ {3 + \cos \left( {\frac{{3\pi }}{4} + x} \right) + \cos \left( {\frac{\pi }{4} + x} \right) + \cos \left( {\frac{\pi }{4} - x} \right) - \cos \left( {\frac{{3\pi }}{4} - x} \right)} \right]$

The range of the function $f(x) = {\log _{\sqrt 5 }}\left[ {3 + \cos \left( {\frac{{3\pi }}{4} + x} \right) + \cos \left( {\frac{\pi }{4} + x} \right) + \cos \left( {\frac{\pi }{4} - x} \right) - \cos \left( {\frac{{3\pi }}{4} - x} \right)} \right]$ is: (A) $[ - 2,2]$ (B) $\left[ {\frac{1}{{\sqrt 5 }},\sqrt 5 } \right]$ (C) $(0,\sqrt 5 )$ (D) $[ 0,2]$ Solution We have, $f(x) = {\log _{\sqrt 5 }}\left( {3 - 2\sin \frac{{3\pi }}{4}\sin x + 2\cos \frac{\pi }{4}\cos x} \right)$ $ \Rightarrow f(x) = {\log _{\sqrt 5 }}\left[ {3 + \sqrt 2 (\cos x - \sin x)} \right]$ Now, $ - \sqrt 2  \le \cos x - \sin x \le \sqrt 2 $ $\therefore - 2 \le \sqrt 2 (\cos x - \sin x) \le 2$ $\therefore 1 \le 3 + \sqrt 2 (\cos x - \sin x) \le 5$ $\therefore{\log _{\sqrt 5 }}1 \le {\log _{\sqrt 5 }}[3 + \sqrt 2 (\cos x - \sin x)] \le {\log _{\sqrt 5 }}5$ $ \Rightarrow 0 \le f(x) \le 2$ Answer: (D)
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