$\int \frac{\cos 2 x . \sin 4 x}{\cos^{4} x (1 + \cos^{2} 2 x)} d x =$

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$\log \frac{1 + \cos^{2} 2 x}{(1 + \cos 2 x)} + s e c x + c$

$\log (\frac{1 + \cos 2 x}{1 + \cos^{2} 2 x}) + s e c^{2} x + c$

$\log (\frac{1 + \cos 2 x}{1 + \cos^{2} x}) + s e c x + c$

$\log \frac{{(1 + \cos 2 x)}^{2}}{(1 + \cos^{2} 2 x)} + s e c^{2} x + c$

answer is C.

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Detailed Solution

$\int \frac{\cos 2 x . \sin 4 x}{\cos^{4} x (1 + \cos^{2} 2 x)} dx = \int \frac{2 \cos^{2} 2 x \sin 2 x}{{(\frac{1 + \cos 2 x}{2})}^{2} (1 + \cos^{2} 2 x)} dx Put \cos 2 x = t - 2 \sin 2 x dx = dt I = \int \frac{- t^{2}}{{(\frac{1 + t}{2})}^{2} (1 + t^{2})} = - 4 \int \frac{t^{2} dt}{{(1 + t)}^{2} (1 + t^{2})} = - 4 \int [\frac{- 1 / 2}{1 + t} + \frac{1 / 2}{{(1 + t)}^{2}} + \frac{\frac{1}{2} t}{1 + t^{2}}] dt (by using partical Fractions)$

$= 2 \log (1 + t) + \frac{2}{1 + t} - \log (1 + t^{2}) + c = \log (\frac{{(1 + t)}^{2}}{1 + t^{2}}) + \frac{2}{1 + t} + c = \log (\frac{{(1 + \cos 2 x)}^{2}}{1 + \cos^{2} 2 x}) + \frac{2}{1 + \cos 2 x} + c = \log (\frac{{(1 + \cos 2 x)}^{2}}{1 + \cos^{2} 2 x}) + \sec^{2} x + c$