Integral of exponential function

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$\int \left(\csc4x\;\cot4x\right)\;e^{\csc4x} \; dx$

**Answer**

$\int \left(\csc4x\;\cot4x\right)\;e^{\csc4x} \; dx$

Substitute $u= \csc4x$. So, $du =-4\csc4x\;\cot4x\; dx$

And, $-\frac{1}{4}\; du = \csc 4x\;\cot4x\; dx$

Then $\int \left(\csc4x\;\cot4x\right)\;e^{\csc4x} \; dx$

$=-\frac{1}{4}\int e^{u} \; du$

$=-\frac{1}{4}e^u + C $

Substituting back $u= \sec 4x$

$=-\frac{1}{4}e^{\csc 4x} + C$