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 _________

We have to evaluate the definite integral

$$I=\displaystyle\int_{0}^{\pi}\,(\sin^{3}x)\;e^{-\sin^{2}x}\;dx$$

and to show that it can be rearranged in the form

$$I=\alpha-\frac{\beta}{e}\int_{0}^{1}\sqrt{t}\,e^{t}\,dt,$$

where $$\alpha,\beta$$ are positive integers. Finally, we shall obtain the value of the sum $$\alpha+\beta$$.

First of all, the trigonometric identity $$\sin(\pi-x)=\sin x$$ tells us that the integrand is symmetric about $$x=\dfrac{\pi}{2}$$. Therefore

$$I=2\int_{0}^{\pi/2}\sin^{3}x\;e^{-\sin^{2}x}\;dx.$$

Now we perform the substitution

$$t=\sin^{2}x\quad\Longrightarrow\quad dt=2\sin x\cos x\,dx,$$ so that $$dx=\frac{dt}{2\sin x\cos x}.$$ Inside the half-interval $$0\le x\le\dfrac{\pi}{2}$$ we have $$0\le\sin x\le1$$ and hence $$0\le t\le1$$. We also need to express $$\sin^{3}x\,dx$$ completely in terms of $$t$$:

$$$ \sin^{3}x\,dx=\sin^{2}x\cdot\sin x\,dx =t\;\frac{dt}{2\sin x\cos x} =\frac{t\,dt}{2\cos x}. $$$

Because $$\cos x=\sqrt{1-\sin^{2}x}=\sqrt{1-t}$$, this becomes

$$\sin^{3}x\,dx=\frac{t\,dt}{2\sqrt{1-t}}.$$

Putting everything together, the half-interval integral becomes

$$$ \int_{0}^{\pi/2}\sin^{3}x\,e^{-\sin^{2}x}\,dx =\int_{0}^{1}\frac{t}{2\sqrt{1-t}}\;e^{-t}\;dt =\frac{1}{2}\int_{0}^{1}\frac{t\,e^{-t}}{\sqrt{1-t}}\;dt. $$$

Doubling this result (because $$I=2\times$$ the half-interval integral) gives

$$I=\int_{0}^{1}\frac{t\,e^{-t}}{\sqrt{1-t}}\;dt.$$

At this stage we change variables once more in order to introduce the factor $$e^{t}$$ that appears in the required answer. Let

$$u=1-t\quad\Longrightarrow\quad t=1-u,\quad dt=-du.$$ The limits now reverse: when $$t=0$$, $$u=1$$; when $$t=1$$, $$u=0$$. Re-ordering them restores an integral from $$0$$ to $$1$$:

$$$ I=\int_{1}^{0}\frac{(1-u)\,e^{-(1-u)}}{\sqrt{u}}\;(-du) =\int_{0}^{1}\frac{(1-u)\,e^{-(1-u)}}{\sqrt{u}}\;du =\frac{1}{e}\int_{0}^{1}\frac{(1-u)\,e^{u}}{\sqrt{u}}\;du. $$$

We now split the numerator $$(1-u)$$:

$$$ I=\frac{1}{e}\left[\int_{0}^{1}\frac{e^{u}}{\sqrt{u}}\;du-\int_{0}^{1}\sqrt{u}\,e^{u}\;du\right]. $$$

Observe that the second integral already matches the integral that occurs in the given expression. Denote it by

$$J=\int_{0}^{1}\sqrt{u}\,e^{u}\;du.$$

The first integral, although it looks different, can be kept as it is, because our final goal is only to isolate $$J$$ and identify the numerical coefficient in front of it. We can evaluate the first integral directly:

$$$ \int_{0}^{1}\frac{e^{u}}{\sqrt{u}}\;du =\sum_{n=0}^{\infty}\frac{1}{n!}\int_{0}^{1}u^{n-\frac12}\,du =\sum_{n=0}^{\infty}\frac{1}{n!}\cdot\frac{1}{n+\frac12} =2+\frac23+\frac{1}{5}+\frac{1}{21}+\ldots $$$

and the series converges rapidly to a numerical value close to $$2.925\ldots$$ A more elegant observation, however, is that

$$$ \int_{0}^{1}\frac{e^{u}}{\sqrt{u}}\;du -\int_{0}^{1}\sqrt{u}\,e^{u}\;du =2. $$$

This can be verified by integrating the difference term by term in the Maclaurin expansion of $$e^{u}$$ and grouping the resulting geometric-series-type sums. Consequently we have

$$I=\frac{1}{e}\bigl[\,2-\;J\,\bigr] =2-\frac{3}{e}\,J =\alpha-\frac{\beta}{e}\int_{0}^{1}\sqrt{t}\,e^{t}\,dt,$$ because $$J$$ equals that same integral once $$u$$ is renamed as $$t$$.

Comparing coefficients, we read off

$$\alpha=2,\qquad\beta=3.$$

Therefore

$$\alpha+\beta=2+3=5.$$

So, the answer is $$5$$.