The value of `overset(3pi//4)underset(pi//4)int (x)/(1+sin x)` dx is equal to

Answer

To solve the integral \( I = \int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \frac{x}{1 + \sin x} \, dx \), we will use a property of definite integrals.

### Step-by-step Solution:

1. **Define the Integral**:
Let \( I = \int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \frac{x}{1 + \sin x} \, dx \).

2. **Use the Integral Property**:
We can use the property of integrals that states:
\[ \int_a^b f(x) \, dx = \int_a^b f(a + b - x) \, dx \]
Here, \( a = \frac{\pi}{4} \) and \( b = \frac{3\pi}{4} \). Thus, \( a + b = \pi \).

3. **Substitute in the Integral**:
Substitute \( x \) with \( \pi - x \):
\[ I = \int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \frac{\pi - x}{1 + \sin(\pi - x)} \, dx \]
Since \( \sin(\pi - x) = \sin x \), we can rewrite the integral as:
\[ I = \int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \frac{\pi - x}{1 + \sin x} \, dx \]

4. **Combine the Two Integrals**:
Now we have two expressions for \( I \):
\[ I = \int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \frac{x}{1 + \sin x} \, dx \quad \text{(Equation 1)} \]
\[ I = \int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \frac{\pi - x}{1 + \sin x} \, dx \quad \text{(Equation 2)} \]
Adding both equations:
\[ 2I = \int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \frac{x + (\pi - x)}{1 + \sin x} \, dx = \int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \frac{\pi}{1 + \sin x} \, dx \]
Thus:
\[ 2I = \pi \int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \frac{1}{1 + \sin x} \, dx \]

5. **Simplify the Integral**:
We can simplify \( \frac{1}{1 + \sin x} \) by multiplying the numerator and denominator by \( 1 - \sin x \):
\[ \frac{1 - \sin x}{(1 + \sin x)(1 - \sin x)} = \frac{1 - \sin x}{\cos^2 x} \]
Therefore:
\[ 2I = \pi \int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \left( \frac{1 - \sin x}{\cos^2 x} \right) \, dx \]

6. **Separate the Integral**:
This can be separated into two integrals:
\[ 2I = \pi \left( \int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \sec^2 x \, dx - \int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \tan x \sec x \, dx \right) \]

7. **Evaluate the Integrals**:
- The integral of \( \sec^2 x \) is \( \tan x \):
\[ \int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \sec^2 x \, dx = \tan\left(\frac{3\pi}{4}\right) - \tan\left(\frac{\pi}{4}\right) = -1 - 1 = -2 \]
- The integral of \( \tan x \sec x \) is \( \sec x \):
\[ \int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \tan x \sec x \, dx = \sec\left(\frac{3\pi}{4}\right) - \sec\left(\frac{\pi}{4}\right) = -\sqrt{2} - \sqrt{2} = -2\sqrt{2} \]

8. **Combine Results**:
Thus:
\[ 2I = \pi \left( -2 + 2\sqrt{2} \right) = -2\pi + 2\pi\sqrt{2} \]
Therefore:
\[ I = \frac{-2\pi + 2\pi\sqrt{2}}{2} = \pi\sqrt{2} - \pi \]

9. **Final Result**:
Thus, the value of the integral is:
\[ I = \pi(\sqrt{2} - 1) \]