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 down the expressions for \( I_{20} \) and \( I_{19} \)

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

\[ I_{19} = \int_0^{\frac{19}{4}} \frac{\sin(37x)}{\sin x} \, dx = \int_0^{4.75} \frac{\sin(37x)}{\sin x} \, dx \]

### Step 2: Set up the expression for \( I_{20} - I_{19} \)

\[ I_{20} - I_{19} = \int_0^{5} \frac{\sin(39x)}{\sin x} \, dx - \int_0^{4.75} \frac{\sin(37x)}{\sin x} \, dx \]

This can be rewritten as:

\[ I_{20} - I_{19} = \int_{4.75}^{5} \frac{\sin(39x)}{\sin x} \, dx + \int_0^{4.75} \left( \frac{\sin(39x)}{\sin x} - \frac{\sin(37x)}{\sin x} \right) dx \]

### Step 3: Use the sine difference identity

Using the identity \( \sin a - \sin b = 2 \sin\left(\frac{a-b}{2}\right) \cos\left(\frac{a+b}{2}\right) \):

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

Thus, we can express the integral as:

\[ I_{20} - I_{19} = \int_0^{4.75} \frac{2 \sin(x) \cos(38x)}{\sin x} \, dx = 2 \int_0^{4.75} \cos(38x) \, dx \]

### Step 4: Evaluate the integral

Now we evaluate:

\[ \int_0^{4.75} \cos(38x) \, dx \]

The integral of \( \cos(kx) \) is \( \frac{1}{k} \sin(kx) \):

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

Evaluating from 0 to 4.75:

\[ \left[ \frac{1}{38} \sin(38x) \right]_0^{4.75} = \frac{1}{38} \left( \sin(38 \times 4.75) - \sin(0) \right) = \frac{1}{38} \sin(180.5) \]

Since \( \sin(180.5) \) is approximately \( -0.5 \):

\[ \int_0^{4.75} \cos(38x) \, dx \approx \frac{-0.5}{38} \]

### Step 5: Final calculation

Thus:

\[ I_{20} - I_{19} = 2 \cdot \frac{-0.5}{38} = \frac{-1}{38} \]

### Conclusion

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

\[ \boxed{-\frac{1}{38}} \]