Integral of exponential function

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$\int \sin36x\;e^{{\cos}^218x} \; dx$

**Answer**

$\int \sin36x\;e^{{\cos}^218x} \; dx$

Substitute $u= {\cos}^218x$

So, $du =-36\sin18x\;\cos18x\; dx$

Or, $du =-18\sin36x\; dx$

And, $-\frac{1}{18}\; du = \sin36x\; dx$

Then $\int \sin36x\;e^{{\cos}^218x} \; dx$

$=-\frac{1}{18}\int e^{u} \; du$

$=-\frac{1}{18}e^u + C $

Substituting back $u= {\cos}^218x$

$=-\frac{1}{18}e^{{\cos}^2 18x} + C$