If $$\int_0^\pi {({{\sin }^3}x){e^{ - {{\sin }^2}x}}dx = \alpha - {\beta \over e}\int_0^1 {\sqrt t {e^t}dt} } $$, then $\alpha$ + $\beta$ is equal to ____________.

$I = 2\int_0^{\pi /2} {{{\sin }^3}x{e^{ - {{\sin }^2}x}}dx}$<br><br>$$ = 2\int_0^{\pi /2} {\sin x{e^{ - {{\sin }^2}x}}dx} + \int\limits_0^{\pi /2} {\mathop {\cos x}\limits_I \,\underbrace {{e^{ - {{\sin }^2}x}}( - \sin 2x)}_{II}dx} $$$$ = 2\int\limits_0^{\pi /2} {\sin x{e^{ - {{\sin }^2}x}}dx} + \left[ {\cos x{e^{ - {{\sin }^2}x}}} \right]_0^{\pi /2} + \int\limits_0^{\pi /2} {\sin x{e^{ - {{\sin }^2}x}}} dx$$<br><br>$= 3\int\limits_0^{\pi /2} {\sin x{e^{ - {{\sin }^2}x}}dx} - 1$<br><br>$$ = {3 \over 2}\int\limits_{ - 1}^0 {{{{e^\alpha }d\alpha } \over {\sqrt {1 + \alpha } }} - 1} $$ (Put $-$sin²x = t)<br><br>$= {3 \over {2e}}\int\limits_0^1 {{{{e^x}} \over {\sqrt x }}dx - 1}$ (put 1 + $\alpha$ = x)<br><br>$$ = {3 \over {2e}}\int\limits_0^1 {\mathop {{e^x}}\limits_{II} } \mathop {{1 \over {\sqrt x }}}\limits_{II} dx - 1$$<br><br>$= 2 - {3 \over e}\int\limits_0^1 {{e^x}\sqrt x } dx$<br><br>Hence, $\alpha$ + $\beta$ = 5