Integral of exponential function

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$\int \left(\sin3x\right)\;e^{\cos3x} \; dx$

**Answer**

$\int \left(\sin3x\right)\;e^{\cos3x} \; dx$

Substitute $u= \cos3x$. So, $du =-3\sin3x\; dx$

And, $-\frac{1}{3}\; du = \sin 3x\; dx$

Then $\int \left(\sin3x\right)\;e^{\cos3x} \; dx$

$=-\frac{1}{3}\int e^{u} \; du$

$=-\frac{1}{3}e^u + C $

Substituting back $u= \cos 3x$

$=-\frac{1}{3}e^{\cos 3x} + C$