Integral of exponential function

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$\int \left(\sin19x\right)\;e^{\cos19x} \; dx$

**Answer**

$\int \left(\sin19x\right)\;e^{\cos19x} \; dx$

Substitute $u= \cos19x$. So, $du =-19\sin19x\; dx$

And, $-\frac{1}{19}\; du = \sin 19x\; dx$

Then $\int \left(\sin19x\right)\;e^{\cos19x} \; dx$

$=-\frac{1}{19}\int e^{u} \; du$

$=-\frac{1}{19}e^u + C $

Substituting back $u= \cos 19x$

$=-\frac{1}{19}e^{\cos 19x} + C$