Integral of exponential function

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$\int \sin2x\;e^{{\cos}^2x} \; dx$

**Answer**

$\int \sin2x\;e^{{\cos}^2x} \; dx$

Substitute $u= {\cos}^2x$

So, $du =-2\sinx\;\cosx\; dx$

Or, $du =-\sin2x\; dx$

And, $-\; du = \sin2x\; dx$

Then $\int \sin2x\;e^{{\cos}^2x} \; dx$

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

$=-e^u + C $

Substituting back $u= {\cos}^2x$

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