$\int \frac{(\sin x - \cos x)}{(\sin x + \cos x) \sqrt{\sin x \cos x + \sin^{2} x \cos^{2} x}} d x =$

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$\cos e c^{- 1} (1 + 2 \sin x) + c$

$s e c^{- 1} (1 + \sin x) + c$

$- c o s e c^{- 1} (1 + \sin 2 x) + c$

$\cot^{- 1} (1 + \sin x) + c$

answer is C.

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Detailed Solution

$\begin{array}{l} \int \frac{(\sin x - \cos x) d x}{(\sin x + \cos x) \sqrt{\sin x \cos x + \sin^{2} x \cos^{2} x}} \\ = \int \frac{(\sin x - \cos x)}{(\sin x + \cos x) \sqrt{\sin x \cos x (1 + \sin x \cos x)}} d x \\ = \int \frac{(\sin x - \cos x)}{(\sin x + \cos x) \sqrt{\{\frac{(\sin x \cos x)^{2} - 1}{2}\} (1 + \frac{(\sin x + \cos x)^{2} - 1}{2})}} \\ (∴ 2 a b = (a + b)^{2} - (a^{2} + b^{2})) \end{array}$

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

$\begin{array}{l} Let t^{2} = p \\ 2 t d t = d p \\ = - 2 \int \frac{d p}{2 t \cdot t \sqrt{{(t^{2})}^{2} - 1}} \\ = - 2 \int \frac{d p}{2 . t^{2} \sqrt{{(t^{2})}^{2} - 1}} \\ = - \int \frac{d p}{p \sqrt{p^{2} - 1}} \\ = - {cosec}^{- 1} (p) + c \\ = - {cosec}^{- 1} (t^{2}) + c \\ = - {cosec}^{- 1} ((\sin x + \cos x)^{2}) + c \end{array}$