$\int \frac{\sin^{3} x d x}{(\cos^{4} x + 3 \cos^{2} x + 1) \tan^{- 1} (\sec x + \cos x)}$

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$\log |T a n^{- 1} (S e c x + C o s x)| + C$

$\log |C o t^{- 1} (C o s e x + C o t x)| + C$

$\log |T a n^{- 1} (S e c x - T a n x)| + C$

$\log |C o t^{- 1} (C o s e x - C o t x)| + C$

answer is A.

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Detailed Solution

$\begin{array}{l} I = \int \frac{\sin^{3} x d x}{(\cos^{4} x + 3 \cos^{2} x + 1) \tan^{- 1} (\sec x + \cos x)} \\ Put \tan^{- 1} (\sec x + \cos x) = t \\ or \frac{1}{1 + (\sec x + \cos x)^{2}} (\sec x \tan x - \sin x) d x = d t \\ or \frac{\cos^{2} x}{\cos^{2} x + {(1 + \cos^{2} x)}^{2}} \sin x (\frac{1}{\cos^{2} x} - 1) d x = d t \end{array}$

$\begin{array}{l} or \frac{\cos^{2} x \sin x \frac{(1 - \cos^{2} x)}{\cos^{2} x}}{\cos^{2} x + 1 + \cos^{4} x + 2 \cos^{2} x} d x = d t \\ or \frac{\sin^{3} x}{\cos^{4} x + 3 + \cos^{2} x + 1} d x = d t \\ \begin{aligned} I = \int \frac{d t}{t} & = \log | t | + c \\ = \log |\tan^{- 1} (\sec x + \cos x)| + c \end{aligned} \end{array}$