Integral of exponential function

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

**Answer**

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

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

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

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

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

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

Substituting back $u= \cos 17x$

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