Integral of exponential function

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$\int \left(\sin2x\right)\;e^{\cos2x} \; dx$

**Answer**

$\int \left(\sin2x\right)\;e^{\cos2x} \; dx$

Substitute $u= \cos2x$. So, $du =-2\sin2x\; dx$

And, $-\frac{1}{2}\; du = \sin 2x\; dx$

Then $\int \left(\sin2x\right)\;e^{\cos2x} \; dx$

$=-\frac{1}{2}\int e^{u} \; du$

$=-\frac{1}{2}e^u + C $

Substituting back $u= \cos 2x$

$=-\frac{1}{2}e^{\cos 2x} + C$