Question

If $I_n = \int_{0}^{\pi} e^x (\sin x)^n dx$, then $5\frac{I_3}{I_1}$, is equal to __________.

Answer 1

3

Answer 2

To solve this problem, we need to evaluate the given definite integral $I_n = \int_{0}^{\pi} e^x (\sin x)^n dx$ and then find the value of $5\frac{I_3}{I_1}$.

First, let's derive a reduction formula for $I_n$. We will use integration by parts twice. Let $I_n = \int_{0}^{\pi} e^x (\sin x)^n dx$. Apply integration by parts with $u = (\sin x)^n$ and $dv = e^x dx$. Then $du = n (\sin x)^{n-1} (\cos x) dx$ and $v = e^x$.

$I_n = \left[ e^x (\sin x)^n \right]_{0}^{\pi} - \int_{0}^{\pi} e^x n (\sin x)^{n-1} (\cos x) dx$

Evaluate the boundary term: At $x=\pi$: $e^{\pi} (\sin \pi)^n = e^{\pi} (0)^n = 0$ (for $n \ge 1$). At $x=0$: $e^0 (\sin 0)^n = 1 \cdot (0)^n = 0$ (for $n \ge 1$). So, the boundary term $\left[ e^x (\sin x)^n \right]_{0}^{\pi} = 0$.

Therefore, $I_n = - n \int_{0}^{\pi} e^x (\sin x)^{n-1} (\cos x) dx$. (Equation 1)

Now, let's apply integration by parts again to the integral $J = \int_{0}^{\pi} e^x (\sin x)^{n-1} (\cos x) dx$. Let $u' = (\sin x)^{n-1} \cos x$ and $dv' = e^x dx$. Then $v' = e^x$. To find $du'$, we differentiate $u'$ using the product rule: $du' = \left[ (n-1)(\sin x)^{n-2}(\cos x)(\cos x) + (\sin x)^{n-1}(-\sin x) \right] dx$ $du' = \left[ (n-1)(\sin x)^{n-2}\cos^2 x - (\sin x)^n \right] dx$ Substitute $\cos^2 x = 1 - \sin^2 x$: $du' = \left[ (n-1)(\sin x)^{n-2}(1 - \sin^2 x) - (\sin x)^n \right] dx$ $du' = \left[ (n-1)(\sin x)^{n-2} - (n-1)(\sin x)^n - (\sin x)^n \right] dx$ $du' = \left[ (n-1)(\sin x)^{n-2} - n(\sin x)^n \right] dx$.

Now apply integration by parts to $J$: $J = \left[ e^x (\sin x)^{n-1} \cos x \right]_{0}^{\pi} - \int_{0}^{\pi} e^x \left[ (n-1)(\sin x)^{n-2} - n(\sin x)^n \right] dx$.

Evaluate the boundary term $\left[ e^x (\sin x)^{n-1} \cos x \right]_{0}^{\pi}$: At $x=\pi$: $e^{\pi} (\sin \pi)^{n-1} \cos \pi = e^{\pi} (0)^{n-1} (-1) = 0$ (for $n-1 \ge 1$, i.e., $n \ge 2$). At $x=0$: $e^0 (\sin 0)^{n-1} \cos 0 = 1 \cdot (0)^{n-1} \cdot 1 = 0$ (for $n-1 \ge 1$, i.e., $n \ge 2$). So, for $n \ge 2$, the boundary term is $0$.

Thus, for $n \ge 2$: $J = - \int_{0}^{\pi} e^x \left[ (n-1)(\sin x)^{n-2} - n(\sin x)^n \right] dx$ $J = - (n-1) \int_{0}^{\pi} e^x (\sin x)^{n-2} dx + n \int_{0}^{\pi} e^x (\sin x)^n dx$ Recognize $I_{n-2}$ and $I_n$: $J = - (n-1) I_{n-2} + n I_n$.

Substitute this expression for $J$ back into Equation 1: $I_n = - n \left[ -(n-1) I_{n-2} + n I_n \right]$ $I_n = n(n-1) I_{n-2} - n^2 I_n$.

Now, rearrange the terms to solve for $I_n$: $I_n + n^2 I_n = n(n-1) I_{n-2}$ $(1+n^2) I_n = n(n-1) I_{n-2}$ $I_n = \frac{n(n-1)}{1+n^2} I_{n-2}$.

This is the reduction formula for $I_n$. We need to find $5\frac{I_3}{I_1}$. Using the reduction formula for $n=3$: $I_3 = \frac{3(3-1)}{1+3^2} I_{3-2}$ $I_3 = \frac{3 \times 2}{1+9} I_1$ $I_3 = \frac{6}{10} I_1$ $I_3 = \frac{3}{5} I_1$.

Now, we can find the ratio $\frac{I_3}{I_1}$: $\frac{I_3}{I_1} = \frac{3}{5}$.

Finally, calculate $5\frac{I_3}{I_1}$: $5\frac{I_3}{I_1} = 5 \times \frac{3}{5} = 3$.

Note: We could also calculate $I_1$ explicitly, though it's not required for this specific problem. $I_1 = \int_{0}^{\pi} e^x \sin x dx$. Using the standard integral formula $\int e^{ax} \sin(bx) dx = \frac{e^{ax}}{a^2+b^2}(a \sin(bx) - b \cos(bx))$, with $a=1, b=1$: $I_1 = \left[ \frac{e^x}{1^2+1^2}(1 \cdot \sin x - 1 \cdot \cos x) \right]_{0}^{\pi}$ $I_1 = \left[ \frac{e^x}{2}(\sin x - \cos x) \right]_{0}^{\pi}$ $I_1 = \frac{e^{\pi}}{2}(\sin \pi - \cos \pi) - \frac{e^0}{2}(\sin 0 - \cos 0)$ $I_1 = \frac{e^{\pi}}{2}(0 - (-1)) - \frac{1}{2}(0 - 1)$ $I_1 = \frac{e^{\pi}}{2}(1) - \frac{1}{2}(-1)$ $I_1 = \frac{e^{\pi}}{2} + \frac{1}{2} = \frac{e^{\pi}+1}{2}$.

The final answer is $\boxed{3}$.

Explanation of the solution:

1. Define $I_n = \int_{0}^{\pi} e^x (\sin x)^n dx$.
2. Apply integration by parts twice to $I_n$ to derive a recurrence relation.
   • First integration by parts: $I_n = [e^x (\sin x)^n]_0^\pi - n \int_0^\pi e^x (\sin x)^{n-1} \cos x dx$. The boundary term is zero.
   • Let $J = \int_0^\pi e^x (\sin x)^{n-1} \cos x dx$. Apply integration by parts to $J$: $J = [e^x (\sin x)^{n-1} \cos x]_0^\pi - \int_0^\pi e^x [(n-1)(\sin x)^{n-2} - n(\sin x)^n] dx$. The boundary term is zero for $n \ge 2$.
   • Substitute $J$ back into the expression for $I_n$: $I_n = -n [-(n-1)I_{n-2} + nI_n]$.
3. Simplify the recurrence relation: $(1+n^2)I_n = n(n-1)I_{n-2}$, which gives $I_n = \frac{n(n-1)}{1+n^2}I_{n-2}$.
4. Use the recurrence relation for $n=3$: $I_3 = \frac{3(3-1)}{1+3^2}I_1 = \frac{6}{10}I_1 = \frac{3}{5}I_1$.
5. Calculate the required expression: $5\frac{I_3}{I_1} = 5 \times \frac{\frac{3}{5}I_1}{I_1} = 5 \times \frac{3}{5} = 3$.