Let ${\int }_{ }^{ }(\sin 3\theta +\sin \theta )\cos \theta .e^{\sin \theta }.d\theta$$=(A.{\sin }^3\theta +B.{\cos }^2\theta +C.\sin \theta +D.\cos \theta +E).e^{\sin \theta }+K$where K is constant of integration

Find the value of $\frac{B}{A}+\frac{D}{E}$

Let $I={\int }_{ }^{ }2.\sin 2\theta .{\cos }^2\theta \times e^{\sin \theta }d\theta$

$={\int }_{ }^{ }4\sin \theta .{\cos }^3\theta .e^{\sin \theta }.d\theta$

Put $\sin \theta =t,so$

$I={\int }_{ }^{ }4t(1-t^2)e^{t}dt=4{\int }_{ }^{ }\underset{I}{\underbrace{t}}.\underset{II}{\underbrace{e^t}}-4{\int }_{ }^{ }{t}^3.e^{t}dt$

$\therefore I=4I_1-4I_2$

We get, $I=-4{\sin }^3\theta -12{\cos }^2\theta -20\sin \theta +32$

$\therefore A=-4,B=-12,C=-20,D=0,E=32$