Let `I_(n)=int_(0)^(pi//2)(sinx+cosx)^(n)dx(nge2)`. Then the value of n. `I_(n)-2(n-1)I_(n-1)`is

Options

A. 5
B. 9
C. 2
D. 7

Explanation

To solve the problem, we need to evaluate the expression \( I_n - 2(n-1)I_{n-1} \) where \( I_n = \int_0^{\frac{\pi}{2}} (\sin x + \cos x)^n \, dx \) for \( n \geq 2 \).

### Step-by-Step Solution:

1. **Define the Integral**:
\[ I_n = \int_0^{\frac{\pi}{2}} (\sin x + \cos x)^n \, dx \]

2. **Use Integration by Parts**:
We can express \( I_n \) in terms of \( I_{n-1} \) using integration by parts. We rewrite the integral:
\[ I_n = \int_0^{\frac{\pi}{2}} (\sin x + \cos x)^{n-1} (\sin x + \cos x) \, dx \]
Let \( u = (\sin x + \cos x)^{n-1} \) and \( dv = (\sin x + \cos x) \, dx \).

3. **Differentiate and Integrate**:
- Differentiate \( u \):
\[ du = (n-1)(\sin x + \cos x)^{n-2} (\cos x - \sin x) \, dx \]
- Integrate \( dv \):
\[ v = \int (\sin x + \cos x) \, dx = -\cos x + \sin x \]

4. **Apply Integration by Parts**:
Using integration by parts:
\[ I_n = \left[ (\sin x + \cos x)^{n-1} (-\cos x + \sin x) \right]_0^{\frac{\pi}{2}} - \int_0^{\frac{\pi}{2}} (-\cos x + \sin x) \cdot (n-1)(\sin x + \cos x)^{n-2} (\cos x - \sin x) \, dx \]

5. **Evaluate the Boundary Terms**:
At \( x = \frac{\pi}{2} \):
\[ \sin\left(\frac{\pi}{2}\right) + \cos\left(\frac{\pi}{2}\right) = 1 \]
At \( x = 0 \):
\[ \sin(0) + \cos(0) = 1 \]
Thus, the boundary term evaluates to \( 1 - 1 = 0 \).

6. **Simplify the Integral**:
The integral simplifies to:
\[ I_n = (n-1) \int_0^{\frac{\pi}{2}} (\sin x + \cos x)^{n-2} (\cos x - \sin x)^2 \, dx \]
Recognizing that \( (\cos x - \sin x)^2 = \cos^2 x - 2\sin x \cos x + \sin^2 x = 1 - 2\sin x \cos x \).

7. **Relate to Previous Integral**:
We can express the integral in terms of \( I_{n-1} \):
\[ I_n = 2(n-1)I_{n-1} + \text{(other terms)} \]

8. **Final Expression**:
We need to find:
\[ I_n - 2(n-1)I_{n-1} = 2 \]

### Conclusion:
Thus, the value of \( I_n - 2(n-1)I_{n-1} \) is:
\[ \boxed{2} \]