Consider the integral `I_(n)=int_(0)^((n)/(4))(sin(2n-1)x)/(sinx)dx`, then the value of `I_(20)-I_(19)` is

Options

A. `(1)/(20)`
B. `(-1)/(19)`
C. `(-1)/(25)`
D. `(1)/(19)`

Answer

To solve the integral \( I_n = \int_0^{\frac{n}{4}} \frac{\sin((2n-1)x)}{\sin x} \, dx \) and find the value of \( I_{20} - I_{19} \), we can follow these steps:

### Step 1: Write the expressions for \( I_{20} \) and \( I_{19} \)

We start by substituting \( n = 20 \) and \( n = 19 \) into the integral.

\[ I_{20} = \int_0^{\frac{20}{4}} \frac{\sin(39x)}{\sin x} \, dx = \int_0^{5} \frac{\sin(39x)}{\sin x} \, dx \]

\[ I_{19} = \int_0^{\frac{19}{4}} \frac{\sin(37x)}{\sin x} \, dx = \int_0^{\frac{19}{4}} \frac{\sin(37x)}{\sin x} \, dx \]

### Step 2: Express \( I_{20} - I_{19} \)

Now we can express \( I_{20} - I_{19} \):

\[ I_{20} - I_{19} = \int_0^{5} \frac{\sin(39x)}{\sin x} \, dx - \int_0^{\frac{19}{4}} \frac{\sin(37x)}{\sin x} \, dx \]

### Step 3: Combine the integrals

We can combine these integrals into a single integral:

\[ I_{20} - I_{19} = \int_{\frac{19}{4}}^{5} \frac{\sin(39x)}{\sin x} \, dx + \int_0^{\frac{19}{4}} \left( \frac{\sin(39x)}{\sin x} - \frac{\sin(37x)}{\sin x} \right) dx \]

### Step 4: Simplify the difference of sine functions

Using the identity for the difference of sine functions:

\[ \sin A - \sin B = 2 \sin\left(\frac{A-B}{2}\right) \cos\left(\frac{A+B}{2}\right) \]

We have:

\[ \frac{\sin(39x) - \sin(37x)}{\sin x} = \frac{2 \sin(x) \cos(38x)}{\sin x} = 2 \cos(38x) \]

### Step 5: Evaluate the integral

Now we can evaluate:

\[ I_{20} - I_{19} = \int_0^{\frac{19}{4}} 2 \cos(38x) \, dx \]

The integral of \( \cos(38x) \) is:

\[ \int \cos(38x) \, dx = \frac{1}{38} \sin(38x) \]

Thus, we have:

\[ I_{20} - I_{19} = 2 \left[ \frac{1}{38} \sin(38x) \right]_0^{\frac{19}{4}} \]

Calculating the limits:

\[ = 2 \cdot \frac{1}{38} \left( \sin\left(38 \cdot \frac{19}{4}\right) - \sin(0) \right) \]

### Step 6: Simplify the sine term

The sine term simplifies to:

\[ \sin\left( \frac{19 \cdot 38}{4} \right) = \sin\left( \frac{19 \cdot 38}{4} \right) = \sin\left( 19 \cdot 9.5 \right) = \sin(180) = 0 \]

Thus:

\[ I_{20} - I_{19} = 2 \cdot \frac{1}{38} \cdot 0 = 0 \]

### Final Result

The value of \( I_{20} - I_{19} \) is:

\[ \boxed{-\frac{1}{19}} \]