`(sin 2 kx)/(sin x)=2[cos x + cos 3 x + …+cos (2k -1)x]`, then value of `I=int_(0)^(pi//2)(sin 2 k x)/(sin x)cos x dx` is :

Options

A. `-pi//2`
B. 0
C. `pi//2`
D. `pi`

Answer

To solve the given integral \( I = \int_{0}^{\frac{\pi}{2}} \frac{\sin(2kx)}{\sin(x)} \cos(x) \, dx \), we can follow these steps:

### Step 1: Substitute the given expression
We know from the problem statement that:
\[ \frac{\sin(2kx)}{\sin(x)} = 2 \left( \cos(x) + \cos(3x) + \ldots + \cos((2k-1)x) \right) \]
Thus, we can rewrite the integral \( I \) as:
\[ I = \int_{0}^{\frac{\pi}{2}} 2 \left( \cos(x) + \cos(3x) + \ldots + \cos((2k-1)x) \right) \cos(x) \, dx \]

### Step 2: Expand the integral
Expanding the integral gives:
\[ I = 2 \int_{0}^{\frac{\pi}{2}} \left( \cos^2(x) + \cos(x) \cos(3x) + \cos(x) \cos(5x) + \ldots + \cos(x) \cos((2k-1)x) \right) \, dx \]

### Step 3: Use the product-to-sum identities
Using the identity \( 2 \cos(a) \cos(b) = \cos(a+b) + \cos(a-b) \), we can simplify each term:
- For \( \cos(x) \cos(3x) \):
\[ \cos(x) \cos(3x) = \frac{1}{2} (\cos(4x) + \cos(2x)) \]
- Similarly for other terms.

### Step 4: Combine and integrate
Now we can write:
\[ I = 2 \left( \int_{0}^{\frac{\pi}{2}} \cos^2(x) \, dx + \frac{1}{2} \int_{0}^{\frac{\pi}{2}} (\cos(4x) + \cos(2x)) \, dx + \ldots \right) \]

### Step 5: Calculate the integrals
1. The integral of \( \cos^2(x) \):
\[ \int \cos^2(x) \, dx = \frac{1}{2} \left( x + \frac{\sin(2x)}{2} \right) \Big|_{0}^{\frac{\pi}{2}} = \frac{\pi}{4} \]
2. The integral of \( \cos(nx) \) for \( n \neq 0 \):
\[ \int_{0}^{\frac{\pi}{2}} \cos(nx) \, dx = 0 \quad \text{for } n \text{ even} \]

### Step 6: Final result
Thus, combining all the results:
\[ I = 2 \left( \frac{\pi}{4} + 0 + 0 + \ldots \right) = \frac{\pi}{2} \]

### Conclusion
The value of the integral \( I \) is:
\[ \boxed{\frac{\pi}{2}} \]