Integral of exponential function

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$\int \left(\csc6x\;\cot6x\right)\;e^{\csc6x} \; dx$

**Answer**

$\int \left(\csc6x\;\cot6x\right)\;e^{\csc6x} \; dx$

Substitute $u= \csc6x$. So, $du =-6\csc6x\;\cot6x\; dx$

And, $-\frac{1}{6}\; du = \csc 6x\;\cot6x\; dx$

Then $\int \left(\csc6x\;\cot6x\right)\;e^{\csc6x} \; dx$

$=-\frac{1}{6}\int e^{u} \; du$

$=-\frac{1}{6}e^u + C $

Substituting back $u= \sec 6x$

$=-\frac{1}{6}e^{\csc 6x} + C$