Integral of exponential function

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$\int \left(\cos8x\right) \; e^{\sin8x} \; dx$

**Answer**

$\int \left(\cos8x\right) \; e^{\sin8x} \; dx$

Substitute $u= \sin8x$. So, $du =8\cos8x\; dx$

And, $\frac{1}{8}\; du = \cos 8x\; dx$

Then $\int \left(\cos8x\right)\;e^{\sin8x} \; dx$

$=\frac{1}{8}\int e^{u} \; du$

$=\frac{1}{8}e^u + C $

Substituting back $u= \sin 8x$

$=\frac{1}{8}e^{\sin 8x} + C$