If `I_(n)=int_(0)^(pi) e^(x)sin^(n)x " dx then " (I_(3))/(I_(1))` is equal to

Options

A. `(3)/(5)`
B. `(1)/(5)`
C. 1
D. `(2)/(5)`

Explanation

To solve the problem, we need to evaluate the ratio \( \frac{I_3}{I_1} \), where \( I_n = \int_0^{\pi} e^x \sin^n x \, dx \).

### Step 1: Define the Integral
We start with the integral:
\[ I_n = \int_0^{\pi} e^x \sin^n x \, dx \]

### Step 2: Use Integration by Parts
We will use integration by parts to evaluate \( I_n \). We can set:
- \( u = \sin^n x \)
- \( dv = e^x \, dx \)

Then, we differentiate and integrate:
- \( du = n \sin^{n-1} x \cos x \, dx \)
- \( v = e^x \)

Using integration by parts:
\[ I_n = \left[ e^x \sin^n x \right]_0^{\pi} - \int_0^{\pi} e^x n \sin^{n-1} x \cos x \, dx \]

### Step 3: Evaluate the Boundary Terms
Evaluating the boundary terms:
\[ \left[ e^x \sin^n x \right]_0^{\pi} = e^{\pi} \sin^n(\pi) - e^0 \sin^n(0) = e^{\pi} \cdot 0 - 1 \cdot 0 = 0 \]
Thus, we have:
\[ I_n = -n \int_0^{\pi} e^x \sin^{n-1} x \cos x \, dx \]

### Step 4: Apply Integration by Parts Again
Now, we need to integrate \( \int_0^{\pi} e^x \sin^{n-1} x \cos x \, dx \). We can apply integration by parts again:
Let:
- \( u = \sin^{n-1} x \)
- \( dv = e^x \cos x \, dx \)

Then:
- \( du = (n-1) \sin^{n-2} x \cos x \, dx \)
- \( v = e^x \cos x \)

After performing integration by parts, we will eventually find a recursive relationship between \( I_n \), \( I_{n-1} \), and \( I_{n-2} \).

### Step 5: Establish the Recursive Relationship
From the integration by parts, we derive:
\[ I_n = \frac{n(n-1)}{1+n^2} I_{n-2} \]

### Step 6: Calculate \( I_3 \) and \( I_1 \)
Using the recursive relationship:
1. For \( n = 3 \):
\[ I_3 = \frac{3 \cdot 2}{1 + 3^2} I_1 = \frac{6}{10} I_1 = \frac{3}{5} I_1 \]

### Step 7: Find the Ratio
Now we can find the ratio:
\[ \frac{I_3}{I_1} = \frac{3}{5} \]

### Final Answer
Thus, the value of \( \frac{I_3}{I_1} \) is:
\[ \boxed{\frac{3}{5}} \]