The value of `int_(pi//4)^(3pi//4) (x)/(1+sin x)` dx is equal to

Options

A. `(sqrt(2)-1)pi`
B. `(sqrt(2+1))pi`
C. `pi`
D. none of these

Answer

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

\[ \int_a^b f(x) \, dx = \int_a^b f(a + b - x) \, dx \]

### Step 1: Apply the property of definite integrals

Let \( a = \frac{\pi}{4} \) and \( b = \frac{3\pi}{4} \). Then, \( a + b = \frac{\pi}{4} + \frac{3\pi}{4} = \pi \).

Using the property, we have:

\[ I = \int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \frac{x}{1 + \sin x} \, dx = \int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \frac{\pi - x}{1 + \sin(\pi - x)} \, dx \]

### Step 2: Simplify the sine function

Since \( \sin(\pi - x) = \sin x \), we can rewrite the integral:

\[ I = \int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \frac{\pi - x}{1 + \sin x} \, dx \]

### Step 3: Split the integral

Now we can express \( I \) as:

\[ I = \int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \frac{\pi}{1 + \sin x} \, dx - \int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \frac{x}{1 + \sin x} \, dx \]

This gives us:

\[ I = \int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \frac{\pi}{1 + \sin x} \, dx - I \]

### Step 4: Solve for \( I \)

Adding \( I \) to both sides:

\[ 2I = \int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \frac{\pi}{1 + \sin x} \, dx \]

Thus,

\[ I = \frac{1}{2} \int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \frac{\pi}{1 + \sin x} \, dx \]

### Step 5: Evaluate the integral \( \int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \frac{1}{1 + \sin x} \, dx \)

To evaluate this integral, we can multiply the numerator and denominator by \( 1 - \sin x \):

\[ \int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \frac{1 - \sin x}{(1 + \sin x)(1 - \sin x)} \, dx = \int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \frac{1 - \sin x}{\cos^2 x} \, dx \]

This simplifies to:

\[ \int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \sec^2 x - \tan x \sec^2 x \, dx \]

### Step 6: Integrate

The integral of \( \sec^2 x \) is \( \tan x \) and the integral of \( \tan x \sec^2 x \) is \( \frac{1}{2} \tan^2 x \):

\[ \int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \sec^2 x \, dx - \int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \tan x \sec^2 x \, dx \]

Evaluating these integrals gives:

\[ \tan\left(\frac{3\pi}{4}\right) - \tan\left(\frac{\pi}{4}\right) - \frac{1}{2} \left( \tan^2\left(\frac{3\pi}{4}\right) - \tan^2\left(\frac{\pi}{4}\right) \right) \]

### Step 7: Final Calculation

After evaluating the integrals and substituting back, we find:

\[ I = \frac{\pi}{2} - 1 \]

Thus, the final value of the integral is:

\[ \boxed{1} \]