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Answer: $-\frac{1}{5} \cot ^{5} x+\frac{1}{3} \cot ^{3} x-\cot x-x+C$

Hint :- Use substitution method to solve this integral.

Given: $\int \cot ^{6} x d x$

Solution: Let,$I=\int \cot ^{6} x d x$

Re-Write,$\mathrm{I}=\int \cot ^{6} x d x$

$\mathrm{I}=\int \cot ^{2} x \cdot \cot ^{4} x d x$

$\mathrm{I}=\int\left(\operatorname{cosec}^{2} x-1\right) \cot ^{4} x d x$                                                $\text { (if, } \operatorname{cosec}^{2} x-\cot ^{2} x=1 \text { ) }$

\begin{aligned} &I=\int \operatorname{cosec}^{2} x \cdot \cot ^{4} x d x-\int \cot ^{4} x d x \\ &I=\int \operatorname{cosec}^{2} x \cdot \cot ^{4} x d x-\int \cot ^{2} x \cdot \cot ^{2} x d x \end{aligned}

$\inline \left.I=\int \operatorname{cosec}^{2} x \cdot \cot ^{4} x d x-\int \cot ^{2} x\left(\operatorname{cosec}^{2} x-1\right) d x \quad \text { (if, } \operatorname{cosec}^{2} x-\cot ^{2} x=1\right)$

\inline \begin{aligned} &1=\int \operatorname{cosec}^{2} x \cdot \cot ^{4} x d x-\int\left(\cot ^{2} x \cdot \operatorname{cosec}^{2} x-\cot ^{2} x\right) d x \\ &\mathrm{I}=\int \operatorname{cosec}^{2} x \cdot \cot ^{4} x d x-\int \cot ^{2} x \cdot \operatorname{cosec}^{2} x d x+\int \cot ^{2} x d x \end{aligned}

$\left.I=\int \operatorname{cosec}^{2} x \cdot \cot ^{4} x d x-\int \cot ^{2} x \cdot \operatorname{cosec}^{2} x d x+\int\left(\operatorname{cosec}^{2} x-1\right) d x \text { (if, } \cot ^{2} x=\operatorname{cosec}^{2} x-1\right)$

$\mathrm{I}=\int \operatorname{cosec}^{2} x \cdot \cot ^{4} x d x-\int \cot ^{2} x \cdot \operatorname{cosec}^{2} x d x+\int \operatorname{cosec}^{2} x d x-\int 1 d x$

Substitute,  cotx = t

$-\operatorname{cosec}^{2} x d x=d t, \text { then }$

\begin{aligned} &\mathrm{I}=\int t^{4}(-d t)-\int t^{2}(-d t)+\int \operatorname{cosec}^{2} x d \mathrm{x}-\int 1 \mathrm{dx} \\ &=-\int t^{4} d t+\int t^{2} d t+\int \operatorname{cosec}^{2} x d \mathrm{x}-\int x^{0} \mathrm{dx} \end{aligned}

$=-\frac{t^{4+1}}{4+1}+\frac{t^{2+1}}{2+1}-\cot x-\frac{x^{0+1}}{0+1}+C$                                            $\text { (if, } \left.\int x^{n} d x=\frac{x^{n+1}}{n+1}+c \& \int \operatorname{cosec}^{2} x d x=-\cot x+c\right)$

\begin{aligned} &=-\frac{t^{5}}{5}+\frac{t^{3}}{3}-\cot x-x+C \\ &=-\frac{\cot ^{5} x}{5}+\frac{\cot ^{8} x}{3}-\cot x-x+C \end{aligned}                                        $\left ( if,\cot x=t \right )$

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