If $\int_{0}^{\pi}\left(\sin ^{3} x\right) e^{-\sin ^{2} x} d x=\alpha-\frac{\beta}{e} \int_{0}^{1} \sqrt{t} e^{t} d t$, then $\alpha+\beta$
is equal to_____________
$I=2 \int_{0}^{\pi / 2} \sin ^{3} x e^{-\sin ^{2} x} d x$
$=2 \int_{0}^{\pi / 2} \sin x \mathrm{e}^{-\sin ^{2} x} \mathrm{dx}+\int_{0}^{\pi / 2} \cos x \underbrace{\mathrm{e}^{-\sin ^{2} x}(-\sin 2 x)}_{\mathrm{II}} \mathrm{dx}$
$=2 \int_{0}^{\pi / 2} \sin x e^{-\sin ^{2} x} d x+\left[\cos x e^{-\sin ^{2} x}\right]_{0}^{\pi / 2}$
$+\int_{0}^{\pi / 2} \sin x e^{-\sin ^{2} x} d x$
$=3 \int_{0}^{\pi / 2} \sin x e^{-\sin ^{2} x} d x-1$
$=\frac{3}{2} \int_{-1}^{0} \frac{e^{\alpha} d \alpha}{\sqrt{1+\alpha}}-1\left(\right.$ Put $\left.-\sin ^{2} x=t\right)$
$=\frac{3}{2 e} \int_{0}^{1} \frac{e^{x}}{\sqrt{x}} d x-1($ put $1+\alpha=x)$
$=\frac{3}{2 \mathrm{e}} \int_{0}^{1} \mathrm{e}^{\mathrm{x}} \frac{1}{\sqrt{\mathrm{x}}} \mathrm{dx}-1$
$=2-\frac{3}{e} \int_{0}^{1} e^{x} \sqrt{x} d x$
Hence, $\alpha+\beta=5$
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