`(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 integral \( I = \int_0^{\frac{\pi}{2}} \frac{\sin(2kx)}{\sin(x)} \cos(x) \, dx \), we will use the given identity:

\[ \frac{\sin(2kx)}{\sin(x)} = 2 \left( \cos(x) + \cos(3x) + \ldots + \cos((2k-1)x) \right) \]

### Step 1: Substitute the identity into the integral

We can substitute the identity into the integral:

\[ I = \int_0^{\frac{\pi}{2}} 2 \left( \cos(x) + \cos(3x) + \ldots + \cos((2k-1)x) \right) \cos(x) \, dx \]

This simplifies to:

\[ I = 2 \int_0^{\frac{\pi}{2}} \left( \cos^2(x) + \cos(x) \cos(3x) + \ldots + \cos(x) \cos((2k-1)x) \right) \, dx \]

### Step 2: Evaluate each term in the integral

Now we will evaluate each term separately. We know that:

\[ \int_0^{\frac{\pi}{2}} \cos^2(x) \, dx = \frac{\pi}{4} \]

For the other terms, we use the property of integrals of the product of cosines:

\[ \int_0^{\frac{\pi}{2}} \cos(mx) \cos(nx) \, dx = \begin{cases} \frac{\pi}{4} & \text{if } m = n \\ 0 & \text{if } m \neq n \end{cases} \]

### Step 3: Apply the property to the integral

In our case:

- The first term is \( \cos^2(x) \) which gives \( \frac{\pi}{4} \).
- The second term is \( \cos(x) \cos(3x) \) where \( m = 1 \) and \( n = 3 \), thus it evaluates to 0.
- Similarly, \( \cos(x) \cos(5x) \), \( \cos(x) \cos(7x) \), ..., \( \cos(x) \cos((2k-1)x) \) will also evaluate to 0 since \( m \) and \( n \) are not equal.

### Step 4: Combine the results

Thus, the integral simplifies to:

\[ I = 2 \left( \frac{\pi}{4} + 0 + 0 + \ldots + 0 \right) = 2 \cdot \frac{\pi}{4} = \frac{\pi}{2} \]

### Final Result

So, the value of the integral \( I \) is:

\[ I = \pi \]