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$\int {\frac{{\cos x - \cos 2x}}{{1 + 2\cos x}}dx} $=?

The given integral = $\int {\frac{{2\sin \frac{{3x}}{2}\sin \frac{x}{2}}}{{1 + 2 - 4{{\sin }^2}\frac{x}{2}}}dx} $=$\int {\frac{{2\sin \frac{{3x}}{2}\sin \frac{x}{2}}}{{3 - 4{{\sin }^2}\frac{x}{2}}}dx} $

=$ \int {\frac{{2\sin \frac{{3x}}{2}\sin \frac{x}{2}.\sin \frac{x}{2}}}{{3\sin \frac{x}{2} - 4{{\sin }^3}\frac{x}{2}}}dx} $

=$\int {\frac{{2\sin \frac{{3x}}{2}{{\sin }^2}\frac{x}{2}}}{{\sin \frac{{3x}}{2}}}dx} $

=$\int {2{{\sin }^2}\frac{x}{2}dx} $

=$ \int {(1 - \cos x)dx} $

=$ x - \sin x + C$

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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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