$\int \frac{d x}{\sin^{5} x \cos^{5} x}$ is equal to

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$\frac{1}{4} (\tan^{4} x - \cot^{4} x) + 2 (\tan^{2} x - \cot^{2} x) + C$

$\frac{1}{4} (\tan^{4} x - \cot^{4} x) + 2 (\tan^{2} x - \cot^{2} x) + 6 \log | \tan x | + C$

$\frac{1}{4} (\tan^{4} x - \cot^{4} x) + 2 (\cot^{2} x - \tan^{2} x) + 6 \log | \cot x | + C$

$\begin{array}{l} \frac{1}{4} (\cot^{4} x - \tan^{4} x) + 2 (\cot^{2} x - \tan^{2} x) + 6 \log | \tan x | + C \end{array}$

answer is B.

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Detailed Solution

Substitute $\tan x = t, \sec^{2} x d x = d t \Rightarrow d x = \frac{d t}{1 + t^{2}}$

$\int \frac{d x}{\sin^{5} x \cos^{5} x} = \int \frac{d t}{(1 + t^{2}) {(\frac{t}{\sqrt{1 + t^{2}}})}^{5} {(\frac{1}{\sqrt{1 + t^{2}}})}^{5}}$

$\begin{array}{l} = \int \frac{{(1 + t^{2})}^{4}}{t^{5}} d t \\ = \int (\frac{1}{t^{5}} + \frac{4}{t^{3}} + 6 \frac{1}{t} + 4 t + t^{3}) d t \\ = - \frac{t^{- 4}}{4} - 2 t^{- 2} + 6 \log | t | + 4 \frac{t^{2}}{2} + \frac{t^{4}}{4} + C \\ = \frac{1}{4} (\tan^{4} x - \cot^{4} x) + 2 (\tan^{2} x - \cot^{2} x) + 6 \log | \tan x | + C \end{array}$