Question

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 ________.

Hint:

When an integral cannot be solved in elementary terms, but the question relates it to another unsolved integral, the key is often to use substitution and integration by parts to transform one integral's form into the other.

Answer 1

Correct Answer: 5

Let \[ I=\int_0^\pi \sin^3 x\,e^{-\sin^2 x}\,dx \] Rewrite: \[ \sin^3 x = \sin^2 x \sin x \] Substitute: \[ u=\cos x \Rightarrow du=-\sin x\,dx \] When $x=0,\ u=1$ and when $x=\pi,\ u=-1$. \[ I=\int_1^{-1}(1-u^2)e^{-(1-u^2)}(-du) =\frac{1}{e}\int_{-1}^1(1-u^2)e^{u^2}du \] The integrand is even, so: \[ I=\frac{2}{e}\int_0^1(1-u^2)e^{u^2}du =\frac{2}{e}\left(\int_0^1 e^{u^2}du-\int_0^1 u^2e^{u^2}du\right) \] Evaluate $\displaystyle \int_0^1 u^2e^{u^2}du$ using integration by parts: Let \[ \int_0^1 u^2e^{u^2}du=\left[\frac{u}{2}e^{u^2}\right]_0^1-\frac12\int_0^1 e^{u^2}du =\frac{e}{2}-\frac12\int_0^1 e^{u^2}du \] Substitute back: \[ I=\frac{2}{e}\left(\frac32\int_0^1 e^{u^2}du-\frac{e}{2}\right) =\frac{3}{e}\int_0^1 e^{u^2}du-1 \] Now relate to the given integral: \[ \int_0^1\sqrt{t}\,e^t dt \] Let $t=u^2$, then $dt=2u\,du$: \[ \int_0^1\sqrt{t}\,e^t dt =2\int_0^1 u^2e^{u^2}du \] From earlier, \[ \int_0^1 u^2e^{u^2}du=\frac{e}{2}-\frac12\int_0^1 e^{u^2}du \] Hence, \[ \int_0^1 e^{u^2}du=e-\int_0^1\sqrt{t}\,e^t dt \] Substitute into $I$: \[ I=\frac{3}{e}\left(e-\int_0^1\sqrt{t}\,e^t dt\right)-1 =2-\frac{3}{e}\int_0^1\sqrt{t}\,e^t dt \] Comparing with \[ I=\alpha-\frac{\beta}{e}\int_0^1\sqrt{t}\,e^t dt \] we get: \[ \alpha=2,\quad \beta=3 \] \[ \boxed{\alpha+\beta=5} \]