If `int_(0)^(pi//2) cos^(m) x sin^(m) x dx= lamda int_(0)^(pi//2) sin^(m) xdx`, then `lamda`=

Options

A. `2^(m)`
B. `2^(-m)`
C. `sqrt(2^(m))`
D. none

Answer

To solve the given integral equation

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

we will find the value of \(\lambda\).

### Step 1: Rewrite the left-hand side
We start with the left-hand side:

\[ \int_{0}^{\frac{\pi}{2}} \cos^m x \sin^m x \, dx. \]

We can use the identity \(2 \cos x \sin x = \sin(2x)\) to rewrite the integral:

\[ \int_{0}^{\frac{\pi}{2}} \cos^m x \sin^m x \, dx = \frac{1}{2^m} \int_{0}^{\frac{\pi}{2}} \sin(2x)^m \, dx. \]

### Step 2: Change of variable
Now, we will perform a change of variable. Let \(t = 2x\), then \(dt = 2dx\) or \(dx = \frac{dt}{2}\). The limits change as follows: when \(x = 0\), \(t = 0\) and when \(x = \frac{\pi}{2}\), \(t = \pi\). Thus, we have:

\[ \int_{0}^{\frac{\pi}{2}} \cos^m x \sin^m x \, dx = \frac{1}{2^m} \cdot \frac{1}{2} \int_{0}^{\pi} \sin^m t \, dt = \frac{1}{2^{m+1}} \int_{0}^{\pi} \sin^m t \, dt. \]

### Step 3: Use symmetry of the sine function
Using the property of the sine function, we know:

\[ \int_{0}^{\pi} \sin^m t \, dt = 2 \int_{0}^{\frac{\pi}{2}} \sin^m t \, dt. \]

So we can write:

\[ \int_{0}^{\frac{\pi}{2}} \cos^m x \sin^m x \, dx = \frac{1}{2^{m+1}} \cdot 2 \int_{0}^{\frac{\pi}{2}} \sin^m t \, dt = \frac{1}{2^m} \int_{0}^{\frac{\pi}{2}} \sin^m t \, dt. \]

### Step 4: Set the equation
Now we can substitute this back into our original equation:

\[ \frac{1}{2^m} \int_{0}^{\frac{\pi}{2}} \sin^m t \, dt = \lambda \int_{0}^{\frac{\pi}{2}} \sin^m t \, dt. \]

### Step 5: Solve for \(\lambda\)
Assuming \(\int_{0}^{\frac{\pi}{2}} \sin^m t \, dt \neq 0\), we can divide both sides by \(\int_{0}^{\frac{\pi}{2}} \sin^m t \, dt\):

\[ \frac{1}{2^m} = \lambda. \]

Thus, we find:

\[ \lambda = \frac{1}{2^m}. \]

### Final Answer
So, the value of \(\lambda\) is:

\[ \lambda = \frac{1}{2^m}. \]