Integral of exponential function

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$\int \sin8x\;e^{{\cos}^24x} \; dx$

**Answer**

$\int \sin8x\;e^{{\cos}^24x} \; dx$

Substitute $u= {\cos}^24x$

So, $du =-8\sin4x\;\cos4x\; dx$

Or, $du =-4\sin8x\; dx$

And, $-\frac{1}{4}\; du = \sin8x\; dx$

Then $\int \sin8x\;e^{{\cos}^24x} \; dx$

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

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

Substituting back $u= {\cos}^24x$

$=-\frac{1}{4}e^{{\cos}^2 4x} + C$