If ${\int }_0^{\pi }\left({\sin }^{3}x\right)e^{-{\sin }^{2}x}dx=\alpha -\frac{\beta }{e}{\int }_0^1\sqrt{t}{e}^{t}dt,$ then $\alpha +\beta$ is equal to

Let $I={\int }_0^{\pi }{\sin }^{3}x{e}^{-{\sin }^{2}x}dx$

$\Rightarrow I=2{\int }_0^{\frac{\pi }{2}}{\sin }^{3}x{e}^{-{\sin }^{2}x}dx \left\{\text{applying} {\int }_0^{a}f\left(x\right)=2{\int }_0^{\frac{a}{2}}f\left(x\right) \text{when} f\left(x\right)=f\left(a-x\right)\right\}$

$=2{\int }_0^{\frac{\mathrm{\pi }}{2}}\sin x\left(1-{\cos }^{2}x\right)e^{-{\sin }^{2}x}dx$

$=2{\int }_0^{\frac{\mathrm{\pi }}{2}}\sin x{e}^{-{\sin }^{2}x}dx-{\int }_0^{\frac{\mathrm{\pi }}{2}}2\sin x\cos x·\cos x{e}^{-{\sin }^{2}x}dx$

$=2{\int }_0^{\frac{\pi }{2}}\sin x{\mathrm{e}}^{-{\sin }^{2}x} dx+{\int }_0^{\frac{\pi }{2}}\underset{I}{\underbrace{\cos x}}·\underset{\mathrm{II}}{\underbrace{{\mathrm{e}}^{-{\sin }^{2}x}\left(-\sin 2x\right)}}dx$

$=2{\int }_0^{\frac{\pi }{2}}\sin x{e}^{-{\sin }^{2}x}dx+{\left[\cos x{e}^{-{\sin }^{2}x}\right]}_0^{\frac{\pi }{2}}+{\int }_0^{\frac{\pi }{2}}\sin x{\mathrm{e}}^{-{\sin }^{2}x}dx$

$=3{\int }_0^{\frac{\pi }{2}}\sin x{e}^{-{\sin }^{2}x}dx-1$

$=\frac{3}{2}{\int }_{-1}^0\frac{e^{\alpha }d\alpha }{\sqrt{1+\alpha }}-1 \left(\mathrm{Put}-{\sin }^{2}x=\alpha \right)$

$=\frac{3}{2e}{\int }_0^1\frac{e^x}{\sqrt{x}}dx-1\left(\mathrm{Put} 1+\alpha =x\right)$

$=\frac{3}{2e}{\int }_0^1{e}^x\frac{1}{\sqrt{x}}dx-1=\frac{3}{2e}\left({\left(2\sqrt{x}{e}^x\right)}_0^1-{\int }_0^{1}2\sqrt{x}{e}^{x}dx\right)-1$

$=\frac{3}{2e}\left(\left(2e\right)-{\int }_0^{1}2\sqrt{x}{e}^{x}dx\right)-1$

$=2-\frac{3}{e}{\int }_0^1{e}^x\sqrt{x}dx$

Hence, $\alpha +\beta =5.$