Integral of exponential function

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$\int \left(\csc18x\;\cot18x\right)\;e^{\csc18x} \; dx$

**Answer**

$\int \left(\csc18x\;\cot18x\right)\;e^{\csc18x} \; dx$

Substitute $u= \csc18x$. So, $du =-18\csc18x\;\cot18x\; dx$

And, $-\frac{1}{18}\; du = \csc 18x\;\cot18x\; dx$

Then $\int \left(\csc18x\;\cot18x\right)\;e^{\csc18x} \; dx$

$=-\frac{1}{18}\int e^{u} \; du$

$=-\frac{1}{18}e^u + C $

Substituting back $u= \sec 18x$

$=-\frac{1}{18}e^{\csc 18x} + C$