If `int_(20)^(40)(sinx)/(sinx+sin(60-x))dx=k`, then the value of `(k)/(4)` is equal to

Answer

To solve the integral \( k = \int_{20}^{40} \frac{\sin x}{\sin x + \sin(60 - x)} \, dx \), we can use a property of definite integrals.

### Step-by-step Solution:

1. **Identify the Integral and its Limits**:
We have the integral:
\[ k = \int_{20}^{40} \frac{\sin x}{\sin x + \sin(60 - x)} \, dx \]

2. **Use the Property of Integration**:
We can use the property:
\[ \int_a^b f(x) \, dx = \int_a^b f(a + b - x) \, dx \]
Here, \( a = 20 \) and \( b = 40 \), so \( a + b = 60 \). Thus, we can write:
\[ k = \int_{20}^{40} \frac{\sin(60 - x)}{\sin(60 - x) + \sin x} \, dx \]

3. **Set Up the New Integral**:
Let’s denote the new integral as:
\[ k = \int_{20}^{40} \frac{\sin(60 - x)}{\sin(60 - x) + \sin x} \, dx \]

4. **Add the Two Integrals**:
Now, we have two expressions for \( k \):
\[ k = \int_{20}^{40} \frac{\sin x}{\sin x + \sin(60 - x)} \, dx \]
and
\[ k = \int_{20}^{40} \frac{\sin(60 - x)}{\sin(60 - x) + \sin x} \, dx \]
Adding these two integrals gives:
\[ 2k = \int_{20}^{40} \left( \frac{\sin x}{\sin x + \sin(60 - x)} + \frac{\sin(60 - x)}{\sin(60 - x) + \sin x} \right) \, dx \]

5. **Simplify the Expression**:
The sum inside the integral simplifies to:
\[ \frac{\sin x + \sin(60 - x)}{\sin x + \sin(60 - x)} = 1 \]
Therefore, we have:
\[ 2k = \int_{20}^{40} 1 \, dx \]

6. **Evaluate the Integral**:
Now, evaluate the integral:
\[ \int_{20}^{40} 1 \, dx = [x]_{20}^{40} = 40 - 20 = 20 \]
Thus, we have:
\[ 2k = 20 \]

7. **Solve for \( k \)**:
Dividing both sides by 2 gives:
\[ k = 10 \]

8. **Find \( \frac{k}{4} \)**:
Finally, we need to find \( \frac{k}{4} \):
\[ \frac{k}{4} = \frac{10}{4} = 2.5 \]

### Final Answer:
\[ \frac{k}{4} = 2.5 \]