Integral of exponential function

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$\int \left(\csc^218x\;\cot18x\right)\;e^{{\cot}^218x} \; dx$

**Answer**

$\int \left(\csc^218x\;\cot18x\right)\;e^{{\cot}^218x} \; dx$

Substitute $u= {\cot}^218x$. So, $du =-36{\csc}^218x\;\cot18x\; dx$

And, $-\frac{1}{36}\; du = {\csc}^2 18x\tan18x\; dx$

Then $\int \left({\csc}^218x\;\cot18x\right)\;e^{{\cot}^218x} \; dx$

$=-\frac{1}{36}\int e^{u} \; du$

$=-\frac{1}{36}e^u + C $

Substituting back $u= {\cot}^2 18x$

$=-\frac{1}{36}e^{{\cot}^218x} + C$