Integral of exponential function

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$\int \left({\csc}^26x\;\cot6x\right)\;e^{{\csc}^26x} \; dx$

**Answer**

$\int \left({\csc}^26x\;\cot6x\right)\;e^{{\csc}^26x} \; dx$

Substitute $u= {\csc}^26x$. So, $du =-12{\csc}^26x\cot6x\; dx$

And, $-\frac{1}{12}\; du = {\csc}^2 6x\;\cot6x\; dx$

Then $\int \left({\csc}^26x\;\tan6x\right)\;e^{{\csc}^26x} \; dx$

$=-\frac{1}{12}\int e^{u} \; du$

$=-\frac{1}{12}e^u + C $

Substituting back $u= {\csc}^2 6x$

$=-\frac{1}{12}e^{{\csc}^2 6x} + C$