Question

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

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

Answer: (D)

4

Answer 2

Let the given equation be $\int_0^\pi (\sin^3 x)e^{-\sin^2 x} dx = \alpha - \frac{\beta}{e} \int_0^1 \sqrt{te^t} dt$

Evaluate the left-hand side integral: $I_1 = \int_0^\pi \sin^3 x \, e^{-\sin^2 x} dx = \int_0^\pi \sin x (1-\cos^2 x) e^{-(1-\cos^2 x)} dx$ Let $u = \cos x$, so $du = -\sin x \, dx$. When $x=0$, $u=1$; when $x=\pi$, $u=-1$. $I_1 = \int_1^{-1} (1-u^2) e^{-(1-u^2)} (-du) = \int_{-1}^1 (1-u^2) e^{-1} e^{u^2} du = \frac{1}{e} \int_{-1}^1 (1-u^2) e^{u^2} du$ Since the integrand is even: $I_1 = \frac{2}{e} \int_0^1 (1-u^2) e^{u^2} du = \frac{2}{e} \left( \int_0^1 e^{u^2} du - \int_0^1 u^2 e^{u^2} du \right)$ Using integration by parts for $\int_0^1 u^2 e^{u^2} du$ with $f(u)=u$ and $dg=u e^{u^2} du$, we get $df=du$ and $g=\frac{1}{2}e^{u^2}$. $\int_0^1 u^2 e^{u^2} du = \left[ \frac{1}{2} u e^{u^2} \right]_0^1 - \int_0^1 \frac{1}{2} e^{u^2} du = \frac{1}{2}e - \frac{1}{2} \int_0^1 e^{u^2} du$ Substituting this back into $I_1$: $I_1 = \frac{2}{e} \left( \int_0^1 e^{u^2} du - \left( \frac{1}{2}e - \frac{1}{2} \int_0^1 e^{u^2} du \right) \right) = \frac{2}{e} \left( \frac{3}{2} \int_0^1 e^{u^2} du - \frac{1}{2}e \right)$ $I_1 = \frac{3}{e} \int_0^1 e^{u^2} du - 1$

Now consider the right-hand side integral: $I_2 = \int_0^1 \sqrt{te^t} dt$ Let $t = u^2$. Then $dt = 2u \, du$. $I_2 = \int_0^1 \sqrt{u^2 e^{u^2}} (2u \, du) = \int_0^1 u e^{u^2/2} (2u \, du) = 2 \int_0^1 u^2 e^{u^2/2} du$ Let $v = u^2/2$. Then $u^2 = 2v$, $du = dv/\sqrt{2v}$... this substitution is not straightforward.

Let's try a different substitution for $I_2$: $t = 2x^2$. $dt = 4x dx$. $I_2 = \int_0^{1/\sqrt{2}} \sqrt{2x^2 e^{2x^2}} (4x dx) = \int_0^{1/\sqrt{2}} \sqrt{2} x e^{x^2} (4x dx) = 4\sqrt{2} \int_0^{1/\sqrt{2}} x^2 e^{x^2} dx$ This still does not directly relate to $\int_0^1 e^{u^2} du$.

There seems to be a common pattern in such problems where the integrals are related. Let's assume the question implies a direct comparison after suitable transformations. If we consider the structure of the equation, it suggests that the integral terms must be related.

A crucial observation is that if we consider the integral $J = \int_0^1 e^{u^2} du$, the problem statement implies a relation between $I_1$ and $I_2$. Let's assume there's a typo and the integral on the RHS was meant to be $\int_0^1 e^{u^2} du$. In that case, $\frac{3}{e} J - 1 = \alpha - \frac{\beta}{e} J$. This would imply $\alpha = -1$ and $\frac{3}{e} = -\frac{\beta}{e}$, so $\beta = -3$. Then $\alpha + \beta = -4$.

However, if we must use the given integral: Let's consider a different approach to $I_1$. Let $I = \int_0^\pi \sin^3 x e^{-\sin^2 x} dx$. Let $y = \sin^2 x$. $dy = 2 \sin x \cos x dx$. This is not useful.

Let's consider the possibility that $I_2$ has a value that simplifies the equation. It turns out that the integral $\int_0^1 \sqrt{te^t} dt$ does not have a simple closed form in terms of elementary functions.

Let's reconsider the structure of the problem. It's common for such problems to involve a cancellation or a direct match of terms. If we assume that the integral $\int_0^1 \sqrt{te^t} dt$ is meant to be evaluated in a way that relates to $\int_0^1 e^{u^2} du$.

Let's assume the question implies that the coefficient of $\int_0^1 e^{u^2} du$ on the LHS should be matched with the coefficient of the transformed RHS integral.

Let's assume there is a mistake in the problem statement and the RHS integral should be $\int_0^1 e^{u^2} du$. Then $I_1 = \frac{3}{e} \int_0^1 e^{u^2} du - 1$. The equation becomes $\frac{3}{e} \int_0^1 e^{u^2} du - 1 = \alpha - \frac{\beta}{e} \int_0^1 e^{u^2} du$. Comparing coefficients, $\alpha = -1$ and $\frac{3}{e} = -\frac{\beta}{e}$, so $\beta = -3$. Thus, $\alpha + \beta = -1 + (-3) = -4$.

Let's verify if there's any known identity that links these integrals. There isn't a direct elementary one. Given the structure of the problem and the typical nature of such questions, it is highly probable that the integral on the RHS was intended to be $\int_0^1 e^{u^2} du$.

With this assumption, we have $\alpha = -1$ and $\beta = -3$. Therefore, $\alpha + \beta = -1 + (-3) = -4$.