Integral of exponential function

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$\int \left({\csc}^2x\;\cotx\right)\;e^{{\csc}^2x} \; dx$

**Answer**

$\int \left({\csc}^2x\;\cotx\right)\;e^{{\csc}^2x} \; dx$

Substitute $u= {\csc}^2x$. So, $du =-2{\csc}^2x\cotx\; dx$

And, $-\frac{1}{2}\; du = {\csc}^2 x\;\cotx\; dx$

Then $\int \left({\csc}^2x\;\tanx\right)\;e^{{\csc}^2x} \; dx$

$=-\frac{1}{2}\int e^{u} \; du$

$=-\frac{1}{2}e^u + C $

Substituting back $u= {\csc}^2 x$

$=-\frac{1}{2}e^{{\csc}^2 x} + C$