If the integral `I_(n)=int_(0)^((pi)/(2))(sin(2n-1)x)/(sinx)dx`,. Then the value of `[I_(20)]^(3)-[I_(19)]^(3)`is

Options

A. 400
B. 200
C. 361
D. 0

Answer

To solve the integral \( I_n = \int_0^{\frac{\pi}{2}} \frac{\sin((2n-1)x)}{\sin x} \, dx \), we will first evaluate \( I_{20} \) and \( I_{19} \) and then find the value of \( I_{20}^3 - I_{19}^3 \).

### Step 1: Evaluate \( I_1 \)

For \( n = 1 \):
\[ I_1 = \int_0^{\frac{\pi}{2}} \frac{\sin(1 \cdot x)}{\sin x} \, dx = \int_0^{\frac{\pi}{2}} 1 \, dx = \left[ x \right]_0^{\frac{\pi}{2}} = \frac{\pi}{2} \]

**Hint:** When \( n = 1 \), the integral simplifies to the integral of 1 over the interval.

### Step 2: Evaluate \( I_2 \)

For \( n = 2 \):
\[ I_2 = \int_0^{\frac{\pi}{2}} \frac{\sin(3x)}{\sin x} \, dx \]
Using the identity \( \sin(3x) = 3\sin x - 4\sin^3 x \):
\[ I_2 = \int_0^{\frac{\pi}{2}} \frac{3\sin x - 4\sin^3 x}{\sin x} \, dx = \int_0^{\frac{\pi}{2}} (3 - 4\sin^2 x) \, dx \]
This can be split into two integrals:
\[ I_2 = \int_0^{\frac{\pi}{2}} 3 \, dx - 4 \int_0^{\frac{\pi}{2}} \sin^2 x \, dx \]
Calculating the first integral:
\[ \int_0^{\frac{\pi}{2}} 3 \, dx = 3 \cdot \frac{\pi}{2} = \frac{3\pi}{2} \]
Now, for the second integral, we know:
\[ \int_0^{\frac{\pi}{2}} \sin^2 x \, dx = \frac{\pi}{4} \]
Thus,
\[ I_2 = \frac{3\pi}{2} - 4 \cdot \frac{\pi}{4} = \frac{3\pi}{2} - \pi = \frac{\pi}{2} \]

**Hint:** Use trigonometric identities to simplify the integral.

### Step 3: Generalize \( I_n \)

From the calculations of \( I_1 \) and \( I_2 \), we observe:
\[ I_1 = I_2 = \frac{\pi}{2} \]
We can conjecture that \( I_n = \frac{\pi}{2} \) for all \( n \).

### Step 4: Evaluate \( I_{20} \) and \( I_{19} \)

Since we have established that:
\[ I_{20} = \frac{\pi}{2} \quad \text{and} \quad I_{19} = \frac{\pi}{2} \]

### Step 5: Calculate \( I_{20}^3 - I_{19}^3 \)

Now we can compute:
\[ I_{20}^3 - I_{19}^3 = \left(\frac{\pi}{2}\right)^3 - \left(\frac{\pi}{2}\right)^3 = 0 \]

### Final Answer

Thus, the value of \( I_{20}^3 - I_{19}^3 \) is:
\[ \boxed{0} \]