$\int \frac{d x}{\sin x (2 + \cos x - 2 \sin x)}$ is equal to

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$\ln | \tan \frac{x}{2} | + \ln | \tan \frac{x}{2} + 1 | + c$

$\ln | \tan \frac{x}{2} | + \frac{5}{3} \ln | \tan \frac{x}{2} - 3 | - \ln | \tan \frac{x}{2} - 1 | + c$

$\frac{1}{3} \ln | \tan \frac{x}{2} | + \frac{5}{3} \ln | \tan \frac{x}{2} - 3 | - \ln | \tan \frac{x}{2} - 1 | + c$

$\frac{1}{3} \ln | \tan \frac{x}{2} | + \frac{5}{3} \ln | \tan \frac{x}{2} - 3 | + \ln | \tan \frac{x}{2} - 1 | + c$

answer is C.

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Detailed Solution

Let $I = \int \frac{d x}{\sin x (2 + \cos x - 2 \sin x)}$
Let $\tan \frac{x}{2} = t$
$\Rightarrow \frac{1}{2} \sec^{2} \frac{x}{2} d x = d t$

$\Rightarrow d x = \frac{2 d t}{1 + t^{2}}$
$∴ I = \int \frac{2 d t}{2 t} (2 + \frac{1 - t^{2}}{1 + t^{2}} - \frac{4 t}{1 + t^{2}}) = \int \frac{(1 + t^{2}) d t}{t (t - 3) (t - 1)}$
$∴ I = \frac{1}{3} \ln | \tan \frac{x}{2} | + \frac{5}{3} \ln | \tan \frac{x}{2} - 3 | - \ln | \tan \frac{x}{2} - 1 | + c$