Integral of exponential function

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$\int \left(\cos17x\right) \; e^{\sin17x} \; dx$

**Answer**

$\int \left(\cos17x\right) \; e^{\sin17x} \; dx$

Substitute $u= \sin17x$. So, $du =17\cos17x\; dx$

And, $\frac{1}{17}\; du = \cos 17x\; dx$

Then $\int \left(\cos17x\right)\;e^{\sin17x} \; dx$

$=\frac{1}{17}\int e^{u} \; du$

$=\frac{1}{17}e^u + C $

Substituting back $u= \sin 17x$

$=\frac{1}{17}e^{\sin 17x} + C$