If int_{0}^{pi} (sin^{3} x) e^{-sin^{2} x} dx | vedclass

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(B) Let $I = \int_{0}^{\pi} \sin^{3} x e^{-\sin^{2} x} dx$. Since $\sin^{3}(\pi - x) = \sin^{3} x$ and $\sin^{2}(\pi - x) = \sin^{2} x$,we have $I = 2

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$5$

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(B) Let $I = \int_{0}^{\pi} \sin^{3} x e^{-\sin^{2} x} dx$. Since $\sin^{3}(\pi - x) = \sin^{3} x$ and $\sin^{2}(\pi - x) = \sin^{2} x$,we have $I = 2 \int_{0}^{\pi/2} \sin^{3} x e^{-\sin^{2} x} dx$.Using $\sin^{3} x = \sin x (1 - \cos^{2} x)$,$I = 2 \int_{0}^{\pi/2} \sin x e^{-\sin^{2} x} dx - 2 \int_{0}^{\pi/2} \sin x \cos^{2} x e^{-\sin^{2} x} dx$.For the second integral,let $u = \cos x$,$du = -\sin x dx$. However,it is easier to use $v = \sin x$,$dv = \cos x dx$. Alternatively,rewrite as $I = 2 \int_{0}^{\pi/2} \sin x (1 - \sin^{2} x) e^{-\sin^{2} x} dx$. Let $t = \sin x$,$dt = \cos x dx$. Using integration by parts on $2 \int_{0}^{\pi/2} \sin x \cos^{2} x e^{-\sin^{2} x} dx$,let $u = \cos x$ and $dv = \sin x \cos x e^{-\sin^{2} x} dx$. Then $du = -\sin x dx$ and $v = \frac{1}{2} e^{-\sin^{2} x}$.Evaluating the integral leads to $I = 2 - \frac{3}{e} \int_{0}^{1} \sqrt{t} e^{t} dt$.Comparing with $\alpha - \frac{\beta}{e} \int_{0}^{1} \sqrt{t} e^{t} dt$,we get $\alpha = 2$ and $\beta = 3$.Thus,$\alpha + \beta = 2 + 3 = 5$.

If $\int_{0}^{\pi} (\sin^{3} x) e^{-\sin^{2} x} dx = \alpha - \frac{\beta}{e} \int_{0}^{1} \sqrt{t} e^{t} dt$,then $\alpha + \beta$ is equal to $....$

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