If `I_n = int_0^(pi//2) x^(n) sin x dx and n gt 1` then `I_n + n (n-1) I_(n-2)` is equal to

Options

A. `n(pi/2)^n`
B. `(n-1)(pi/2)^n`
C. `n(pi/2)^(n-1)`
D. `(n-1)(pi/2)^(n-1)`

Answer

To solve the integral \( I_n = \int_0^{\frac{\pi}{2}} x^n \sin x \, dx \) and find the value of \( I_n + n(n-1)I_{n-2} \), we will use integration by parts.

### Step-by-Step Solution:

1. **Integration by Parts Setup**:
We will use the integration by parts formula:
\[ \int u \, dv = uv - \int v \, du \]
Here, we choose:
- \( u = x^n \) (thus \( du = n x^{n-1} \, dx \))
- \( dv = \sin x \, dx \) (thus \( v = -\cos x \))

2. **Applying Integration by Parts**:
Now we apply the integration by parts:
\[ I_n = \int_0^{\frac{\pi}{2}} x^n \sin x \, dx = \left[-x^n \cos x \right]_0^{\frac{\pi}{2}} + \int_0^{\frac{\pi}{2}} n x^{n-1} \cos x \, dx \]

3. **Evaluating the Boundary Terms**:
Evaluating the first term:
- At \( x = \frac{\pi}{2} \), \( \cos\left(\frac{\pi}{2}\right) = 0 \)
- At \( x = 0 \), \( -0^n \cos(0) = 0 \)
Thus, the boundary term evaluates to \( 0 - 0 = 0 \).

4. **Rewriting the Integral**:
Therefore, we have:
\[ I_n = n \int_0^{\frac{\pi}{2}} x^{n-1} \cos x \, dx \]

5. **Applying Integration by Parts Again**:
Now we need to evaluate \( \int_0^{\frac{\pi}{2}} x^{n-1} \cos x \, dx \) using integration by parts again:
Choose:
- \( u = x^{n-1} \) (thus \( du = (n-1)x^{n-2} \, dx \))
- \( dv = \cos x \, dx \) (thus \( v = \sin x \))

Applying integration by parts:
\[ \int_0^{\frac{\pi}{2}} x^{n-1} \cos x \, dx = \left[ x^{n-1} \sin x \right]_0^{\frac{\pi}{2}} - \int_0^{\frac{\pi}{2}} (n-1)x^{n-2} \sin x \, dx \]

6. **Evaluating the Boundary Terms Again**:
Evaluating the first term:
- At \( x = \frac{\pi}{2} \), \( \sin\left(\frac{\pi}{2}\right) = 1 \) gives \( \left(\frac{\pi}{2}\right)^{n-1} \)
- At \( x = 0 \), \( 0^{n-1} \sin(0) = 0 \)
Thus, the boundary term evaluates to \( \left(\frac{\pi}{2}\right)^{n-1} - 0 = \left(\frac{\pi}{2}\right)^{n-1} \).

7. **Final Expression**:
Therefore, we have:
\[ I_n = n \left(\frac{\pi}{2}\right)^{n-1} - n(n-1) I_{n-2} \]

8. **Finding \( I_n + n(n-1)I_{n-2} \)**:
Now we can find:
\[ I_n + n(n-1)I_{n-2} = n \left(\frac{\pi}{2}\right)^{n-1} \]

### Conclusion:
Thus, the final result is:
\[ I_n + n(n-1)I_{n-2} = n \left(\frac{\pi}{2}\right)^{n-1} \]