Let `I_(n)=int_(0)^(pi//2) sin^(n)x dx, nin N`. Then

Options

A. `I_(n):I_(n-2)=n:(n-1)`
B. `I_(n) gt I_(n-2)`
C. `n(I_(n-2)-I_(n))=I_(n-2)`
D. none of these

Answer

To solve the problem, we need to derive a reduction formula for the integral \( I_n = \int_0^{\frac{\pi}{2}} \sin^n x \, dx \).

### Step-by-Step Solution:

1. **Start with the definition of \( I_n \)**:
\[ I_n = \int_0^{\frac{\pi}{2}} \sin^n x \, dx \]

2. **Use the identity for \( \sin^n x \)**:
We can express \( \sin^n x \) as \( \sin^{n-1} x \cdot \sin x \):
\[ I_n = \int_0^{\frac{\pi}{2}} \sin^{n-1} x \cdot \sin x \, dx \]

3. **Apply integration by parts**:
Let \( u = \sin^{n-1} x \) and \( dv = \sin x \, dx \). Then, we have:
\[ du = (n-1) \sin^{n-2} x \cos x \, dx \quad \text{and} \quad v = -\cos x \]
Using integration by parts:
\[ I_n = \left[-\sin^{n-1} x \cos x \right]_0^{\frac{\pi}{2}} + \int_0^{\frac{\pi}{2}} (n-1) \sin^{n-2} x \cos^2 x \, dx \]

4. **Evaluate the boundary terms**:
At \( x = \frac{\pi}{2} \), \( \sin^{n-1} \left(\frac{\pi}{2}\right) = 1 \) and \( \cos \left(\frac{\pi}{2}\right) = 0 \), so the first term evaluates to 0. At \( x = 0 \), \( \sin^{n-1}(0) = 0 \) and \( \cos(0) = 1 \), so the term also evaluates to 0. Thus:
\[ \left[-\sin^{n-1} x \cos x \right]_0^{\frac{\pi}{2}} = 0 \]

5. **Substitute back into the integral**:
Now we have:
\[ I_n = (n-1) \int_0^{\frac{\pi}{2}} \sin^{n-2} x \cos^2 x \, dx \]

6. **Use the identity for \( \cos^2 x \)**:
We know that \( \cos^2 x = 1 - \sin^2 x \):
\[ I_n = (n-1) \left( \int_0^{\frac{\pi}{2}} \sin^{n-2} x \, dx - \int_0^{\frac{\pi}{2}} \sin^{n} x \, dx \right) \]
This can be rewritten as:
\[ I_n = (n-1)(I_{n-2} - I_n) \]

7. **Rearranging the equation**:
Bringing \( I_n \) terms together:
\[ I_n + (n-1) I_n = (n-1) I_{n-2} \]
\[ n I_n = (n-1) I_{n-2} \]

8. **Final reduction formula**:
Thus, we arrive at the reduction formula:
\[ I_n = \frac{(n-1)}{n} I_{n-2} \]