$\int \frac{\sin 2 x + 2 \tan x}{(\cos^{6} x + 6 \cos^{2} x + 4)} dx$

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$\tan^{- 1} \frac{1}{\sqrt{2}} (\frac{1 + \cos^{2} x}{\cos^{7} x}) + c$

$\frac{1}{12} \log_{e} (\frac{1 + \cos^{2} x}{\cos^{7} x}) + c$

$2 \sqrt{\frac{1 + \cos^{2} x}{\cos^{7} x}}$

$\frac{1}{12} \log (1 + \frac{6}{\cos^{4} x} + \frac{4}{\cos^{6} x}) + c$

answer is D.

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Detailed Solution

$I = \int \frac{\sin 2 x + 2 \tan x}{\cos^{6} x + 6 \cos^{2} x + 4} d x \sin c e \sin 2 x + 2 \tan x = 2 \cos x \sin x + \frac{2 \sin x}{\cos x} = (2 \cos x + \frac{2}{\cos x}) \sin x Let \cos x = u \Rightarrow - \sin x d x = d u$
$\begin{array}{l} ∴ I = \int \frac{(- 2 u - \frac{2}{u})}{u^{6} + 6 u^{2} + 4} d u \\ = \int \frac{- 2 u^{- 5} - 2 u^{- 7}}{1 + 6 u^{- 4} + 4 u^{- 6}} d u \end{array}$
$I = \frac{1}{12} \ln |1 + 6 u^{- 4} + 4 u^{- 6}| + c = \frac{1}{12} \ln |1 + 6 \sec^{4} x + 4 \sec^{6} x| + c = \frac{1}{12} \log (1 + \frac{6}{\cos^{4} x} + \frac{4}{\cos^{6} x}) + c$