Integral of exponential function

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$\int \left({\csc}^210x\;\cot10x\right)\;e^{{\csc}^210x} \; dx$

**Answer**

$\int \left({\csc}^210x\;\cot10x\right)\;e^{{\csc}^210x} \; dx$

Substitute $u= {\csc}^210x$. So, $du =-20{\csc}^210x\cot10x\; dx$

And, $-\frac{1}{20}\; du = {\csc}^2 10x\;\cot10x\; dx$

Then $\int \left({\csc}^210x\;\tan10x\right)\;e^{{\csc}^210x} \; dx$

$=-\frac{1}{20}\int e^{u} \; du$

$=-\frac{1}{20}e^u + C $

Substituting back $u= {\csc}^2 10x$

$=-\frac{1}{20}e^{{\csc}^2 10x} + C$