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. **Use the property of definite integrals**:
We know that:
\[ \int_a^b f(x) \, dx = \int_a^b f(a + b - x) \, dx \]
In our case, \( a = 20 \) and \( b = 40 \). Thus, we have:
\[ k = \int_{20}^{40} \frac{\sin x}{\sin x + \sin(60 + x)} \, dx = \int_{20}^{40} \frac{\sin(60 - x)}{\sin(60 - x) + \sin x} \, dx \]

2. **Rewrite the integral**:
We can express the second integral:
\[ k = \int_{20}^{40} \frac{\sin(60 - x)}{\sin(60 - x) + \sin x} \, dx \]

3. **Add the two integrals**:
Now we add the two expressions for \( k \):
\[ 2k = \int_{20}^{40} \left( \frac{\sin x}{\sin x + \sin(60 + x)} + \frac{\sin(60 - x)}{\sin(60 - x) + \sin x} \right) \, dx \]

4. **Simplify the expression**:
The denominators are the same, so we can combine the fractions:
\[ 2k = \int_{20}^{40} \frac{\sin x + \sin(60 - x)}{\sin x + \sin(60 + x)} \, dx \]
Note that \( \sin(60 + x) = \sin(60 - x) \) due to the sine function's properties.

5. **Evaluate the integral**:
The numerator simplifies to \( \sin x + \sin(60 - x) \), and the denominator remains the same. Thus:
\[ 2k = \int_{20}^{40} 1 \, dx \]

6. **Calculate the integral**:
\[ 2k = x \bigg|_{20}^{40} = 40 - 20 = 20 \]
Therefore, we find:
\[ 2k = 20 \implies k = 10 \]

7. **Find \( \frac{k}{4} \)**:
Finally, we compute:
\[ \frac{k}{4} = \frac{10}{4} = 2.5 \]

### Final Answer:
The value of \( \frac{k}{4} \) is \( 2.5 \).