Integral of exponential function

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$\int \left(\sinx\; \cosx\right)\;e^{{\sin}^2x} \; dx$

**Answer**

$\int \left(\sinx\; \cosx\right)\;e^{{\sin}^2x} \; dx$

Substitute $u= {\sin}^2x$. So, $du =2\sinx\;\cosx\; dx$

And, $\frac{1}{2}\; du = \sinx\; \cos x\; dx$

Then $\int \left(\sinx\; \cosx\right)e^{{\sin}^2x} \; dx$

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

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

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

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