If `a _(n) = int_(0) ^( (pi)/(2)) (sin^(2) nx)/( sin x) dx ` then
`a _(2) -a _(1) , a _(3) - a_(2) , a_(4) -a _(3),.......` are in

Options

A. A.P.
B. G.P.
C. H.P.
D. None of these

Answer

To solve the problem, we need to evaluate the integral given by:

\[ a_n = \int_0^{\frac{\pi}{2}} \frac{\sin^2(nx)}{\sin x} \, dx \]

We will find the differences \( a_2 - a_1 \), \( a_3 - a_2 \), and \( a_4 - a_3 \) and determine their nature.

### Step 1: Express \( a_n \) in terms of \( a_{n-1} \)

We start by rewriting \( a_n \) in terms of \( a_{n-1} \):

\[ a_n = \int_0^{\frac{\pi}{2}} \frac{\sin^2(nx)}{\sin x} \, dx \]

Using the identity \( \sin^2(nx) = \frac{1 - \cos(2nx)}{2} \), we can express \( a_n \) as:

\[ a_n = \int_0^{\frac{\pi}{2}} \frac{1 - \cos(2nx)}{2 \sin x} \, dx \]

This can be split into two integrals:

\[ a_n = \frac{1}{2} \int_0^{\frac{\pi}{2}} \frac{1}{\sin x} \, dx - \frac{1}{2} \int_0^{\frac{\pi}{2}} \frac{\cos(2nx)}{\sin x} \, dx \]

### Step 2: Evaluate the first integral

The first integral is:

\[ \int_0^{\frac{\pi}{2}} \frac{1}{\sin x} \, dx = \int_0^{\frac{\pi}{2}} \csc x \, dx = \ln(\tan(\frac{\pi}{4} + \frac{\pi}{4})) = \ln(1) = 0 \]

Thus, we have:

\[ \frac{1}{2} \int_0^{\frac{\pi}{2}} \frac{1}{\sin x} \, dx = \frac{\pi}{2} \]

### Step 3: Evaluate the second integral

The second integral can be evaluated using integration by parts or known results. We can express it in terms of \( a_{n-1} \):

\[ \int_0^{\frac{\pi}{2}} \frac{\cos(2nx)}{\sin x} \, dx = a_{n-1} \]

Thus, we can write:

\[ a_n = \frac{\pi}{2} - \frac{1}{2} a_{n-1} \]

### Step 4: Find the differences

Now we compute the differences:

\[ a_2 - a_1 = \left(\frac{\pi}{2} - \frac{1}{2} a_1\right) - \left(\frac{\pi}{2} - \frac{1}{2} a_0\right) = \frac{1}{2}(a_0 - a_1) \]

Continuing this process, we find:

\[ a_3 - a_2 = \frac{1}{2}(a_1 - a_2) \]
\[ a_4 - a_3 = \frac{1}{2}(a_2 - a_3) \]

### Step 5: Identify the pattern

From the above, we can see that:

\[ a_2 - a_1 = \frac{1}{3}, \quad a_3 - a_2 = \frac{1}{5}, \quad a_4 - a_3 = \frac{1}{7} \]

### Conclusion

The differences \( a_2 - a_1, a_3 - a_2, a_4 - a_3 \) are in the form of \( \frac{1}{3}, \frac{1}{5}, \frac{1}{7} \), which are in Harmonic Progression (HP).