Integral of exponential function

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$\int \left({\csc}^23x\;\cot3x\right)\;e^{{\csc}^23x} \; dx$

**Answer**

$\int \left({\csc}^23x\;\cot3x\right)\;e^{{\csc}^23x} \; dx$

Substitute $u= {\csc}^23x$. So, $du =-6{\csc}^23x\cot3x\; dx$

And, $-\frac{1}{6}\; du = {\csc}^2 3x\;\cot3x\; dx$

Then $\int \left({\csc}^23x\;\tan3x\right)\;e^{{\csc}^23x} \; dx$

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

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

Substituting back $u= {\csc}^2 3x$

$=-\frac{1}{6}e^{{\csc}^2 3x} + C$