Integral of exponential function

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$\int \left(\csc^216x\right)\;e^{\cot16x} \; dx$

**Answer**

$\int \left(\csc^216x\right)\;e^{\cot16x} \; dx$

Substitute $u= \cot16x$. So, $du =-16\csc^216x\; dx$

And, $-\frac{1}{16}\; du = {\csc}^2 16x\; dx$

Then $\int \left(\csc^216x\right)\;e^{\cot16x} \; dx$

$=-\frac{1}{16}\int e^{u} \; du$

$=-\frac{1}{16}e^u + C $

Substituting back $u= \tan 16x$

$=-\frac{1}{16}e^{\cot 16x} + C$