If `int_(0)^(pi//2) cos^(n)x sin^(n) x dx=lambda int_(0)^(pi//2) sin^(n)x" dx"`, then `lambda=`

Options

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

Answer

To solve the given problem, we need to find the value of \(\lambda\) such that:

\[ \int_{0}^{\frac{\pi}{2}} \cos^n x \sin^n x \, dx = \lambda \int_{0}^{\frac{\pi}{2}} \sin^n x \, dx \]

### Step-by-Step Solution:

1. **Start with the Left-Hand Side (LHS)**:
\[ I = \int_{0}^{\frac{\pi}{2}} \cos^n x \sin^n x \, dx \]

2. **Multiply and Divide by \(2^n\)**:
\[ I = \frac{1}{2^n} \int_{0}^{\frac{\pi}{2}} 2^n \cos^n x \sin^n x \, dx \]

3. **Use the Identity \(2 \sin x \cos x = \sin(2x)\)**:
\[ I = \frac{1}{2^n} \int_{0}^{\frac{\pi}{2}} \left(2 \sin x \cos x\right)^n \, dx = \frac{1}{2^n} \int_{0}^{\frac{\pi}{2}} \sin^n(2x) \, dx \]

4. **Change of Variables**:
Let \(t = 2x\), then \(dt = 2dx\) or \(dx = \frac{dt}{2}\). Also, change the limits: when \(x = 0\), \(t = 0\) and when \(x = \frac{\pi}{2}\), \(t = \pi\).
\[ I = \frac{1}{2^n} \cdot \frac{1}{2} \int_{0}^{\pi} \sin^n t \, dt = \frac{1}{2^{n+1}} \int_{0}^{\pi} \sin^n t \, dt \]

5. **Split the Integral**:
The integral from \(0\) to \(\pi\) can be split into two parts:
\[ \int_{0}^{\pi} \sin^n t \, dt = \int_{0}^{\frac{\pi}{2}} \sin^n t \, dt + \int_{\frac{\pi}{2}}^{\pi} \sin^n t \, dt \]
Using the property \(\sin(\pi - t) = \sin t\), we find:
\[ \int_{\frac{\pi}{2}}^{\pi} \sin^n t \, dt = \int_{0}^{\frac{\pi}{2}} \sin^n t \, dt \]
Thus,
\[ \int_{0}^{\pi} \sin^n t \, dt = 2 \int_{0}^{\frac{\pi}{2}} \sin^n t \, dt \]

6. **Substituting Back**:
Now substituting back into our expression for \(I\):
\[ I = \frac{1}{2^{n+1}} \cdot 2 \int_{0}^{\frac{\pi}{2}} \sin^n t \, dt = \frac{1}{2^n} \int_{0}^{\frac{\pi}{2}} \sin^n t \, dt \]

7. **Comparing with the Right-Hand Side (RHS)**:
We have:
\[ I = \lambda \int_{0}^{\frac{\pi}{2}} \sin^n x \, dx \]
Therefore, comparing both sides:
\[ \frac{1}{2^n} \int_{0}^{\frac{\pi}{2}} \sin^n x \, dx = \lambda \int_{0}^{\frac{\pi}{2}} \sin^n x \, dx \]

8. **Solving for \(\lambda\)**:
Dividing both sides by \(\int_{0}^{\frac{\pi}{2}} \sin^n x \, dx\) (assuming it is not zero):
\[ \lambda = \frac{1}{2^n} \]

### Final Answer:
\[ \lambda = \frac{1}{2^n} \]