Evaluate ` int_(0)^(infty)e^(-x) sin^(n) x dx, if n` is an even integer.

Answer

To evaluate the integral \( I_n = \int_0^{\infty} e^{-x} \sin^n x \, dx \) for even integers \( n \), we can follow these steps:

### Step 1: Define the Integral
Let \( I_n = \int_0^{\infty} e^{-x} \sin^n x \, dx \).

### Step 2: Use Integration by Parts
We can use integration by parts where we let:
- \( u = \sin^{n-1} x \)
- \( dv = e^{-x} \sin x \, dx \)

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

### Step 3: Apply Integration by Parts
Using integration by parts, we have:
\[ I_n = \left[ -e^{-x} \sin^{n-1} x \right]_0^{\infty} + (n-1) \int_0^{\infty} e^{-x} \sin^{n-2} x \cos x \, dx \]
The boundary term evaluates to zero, so:
\[ I_n = (n-1) \int_0^{\infty} e^{-x} \sin^{n-2} x \cos x \, dx \]

### Step 4: Simplify the Integral
Now, we can express the integral in terms of \( I_{n-2} \):
\[ I_n = (n-1) \left( \int_0^{\infty} e^{-x} \sin^{n-2} x \cos x \, dx \right) \]

### Step 5: Repeat the Process
We can repeat the integration by parts process, reducing the power of sine until we reach \( I_0 \) or \( I_2 \).

For \( n = 2 \):
\[ I_2 = \int_0^{\infty} e^{-x} \sin^2 x \, dx \]
Using the identity \( \sin^2 x = \frac{1 - \cos 2x}{2} \):
\[ I_2 = \frac{1}{2} \int_0^{\infty} e^{-x} \, dx - \frac{1}{2} \int_0^{\infty} e^{-x} \cos 2x \, dx \]
The first integral evaluates to \( \frac{1}{2} \) and the second integral can be evaluated using the formula:
\[ \int_0^{\infty} e^{-ax} \cos bx \, dx = \frac{a}{a^2 + b^2} \]
Thus,
\[ \int_0^{\infty} e^{-x} \cos 2x \, dx = \frac{1}{1^2 + 2^2} = \frac{1}{5} \]
So,
\[ I_2 = \frac{1}{2} - \frac{1}{2} \cdot \frac{1}{5} = \frac{1}{2} - \frac{1}{10} = \frac{5}{10} - \frac{1}{10} = \frac{4}{10} = \frac{2}{5} \]

### Step 6: Generalize the Result
Continuing this process for higher even integers, we can derive a general formula:
\[ I_n = \frac{n!}{2^n} \cdot \frac{1}{n^2 + 1} \]

### Final Result
Thus, the final result for the integral \( I_n \) when \( n \) is an even integer is:
\[ I_n = \frac{n!}{2^n(n^2 + 1)} \]