Integral of exponential function

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$\int \left(\sin3x\; \cos3x\right)\;e^{{\sin}^23x} \; dx$

**Answer**

$\int \left(\sin3x\; \cos3x\right)\;e^{{\sin}^23x} \; dx$

Substitute $u= {\sin}^23x$. So, $du =6\sin3x\;\cos3x\; dx$

And, $\frac{1}{6}\; du = \sin3x\; \cos 3x\; dx$

Then $\int \left(\sin3x\; \cos3x\right)e^{{\sin}^23x} \; dx$

$=\frac{1}{6}\int e^{u} \; du$

$=\frac{1}{6}e^u + C $

Substituting back $u= {\sin}^23x$

$=\frac{1}{6}e^{{\sin}^2 3x} + C$