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Answer: $-\frac{\cot ^{n+1} x}{n+1}+\mathrm{C}$

Hint:Use substitution method to solve this integral.

Given:$\int \cot ^{n} x \cdot \operatorname{cocec}^{2} x d x, \mathrm{n} \neq-1$

Solution:let,$\mathrm{I}=\int \cot ^{n} x \cdot \operatorname{cocec}^{2} x d x$

Substitute,  cot x = t

$\operatorname{cosec}^{2} x d x=\text { dt then }$

$\begin{array}{ll} \mathrm{I}=\int t^{n} \cdot \operatorname{cosec}^{2} \mathrm{x} \cdot \frac{d t}{-\operatorname{cosec}^{2} x} &\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \text { (if, } \cot \mathrm{x}=\mathrm{t}) \end{array}$

$=-\int t^{n}$

$=-\left[\frac{t^{n+1}}{n+1}\right]+C$                                            $\text { (if, } \left.\int x^{n} d x=\frac{x^{n+1}}{n+1}+c\right)$

$\left.=-\frac{\cot ^{n+1} x}{n+1}+C \: \: \: \: \: \: \: \quad \text { (if, } \mathrm{t}=\cot \mathrm{x}\right)$

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