The value of the integral `int_(-pi/4)^(pi/4)(x+pi/4)/(2-cos2x)dx ` is

Options

A. `pi^2/(12 sqrt3)`
B. `pi^2/(6)`
C. `pi^2/(3 sqrt3)`
D. `pi^2/(6sqrt3)`

Answer

To solve the integral

\[ I = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{x + \frac{\pi}{4}}{2 - \cos 2x} \, dx, \]

we will use the property of definite integrals and the symmetry of the function involved.

### Step 1: Substitute \(x\) with \(-x\)

We start by substituting \(x\) with \(-x\):

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

Since \(\cos(-2x) = \cos(2x)\), we can rewrite the integral as:

\[ I = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{-x + \frac{\pi}{4}}{2 - \cos 2x} \, dx. \]

### Step 2: Combine the two integrals

Now we have two expressions for \(I\):

1. \(I = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{x + \frac{\pi}{4}}{2 - \cos 2x} \, dx\)
2. \(I = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{-x + \frac{\pi}{4}}{2 - \cos 2x} \, dx\)

Adding these two expressions together:

\[ 2I = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \left( \frac{x + \frac{\pi}{4}}{2 - \cos 2x} + \frac{-x + \frac{\pi}{4}}{2 - \cos 2x} \right) \, dx. \]

This simplifies to:

\[ 2I = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{2 \cdot \frac{\pi}{4}}{2 - \cos 2x} \, dx = \frac{\pi}{2} \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{1}{2 - \cos 2x} \, dx. \]

### Step 3: Evaluate the integral

Now we need to evaluate the integral

\[ \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{1}{2 - \cos 2x} \, dx. \]

Using the substitution \(u = 2x\), we have \(du = 2dx\) or \(dx = \frac{du}{2}\). The limits change from \(x = -\frac{\pi}{4}\) to \(x = \frac{\pi}{4}\) which corresponds to \(u = -\frac{\pi}{2}\) to \(u = \frac{\pi}{2}\):

\[ \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{1}{2 - \cos 2x} \, dx = \frac{1}{2} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{1}{2 - \cos u} \, du. \]

### Step 4: Solve the integral

The integral

\[ \int \frac{1}{2 - \cos u} \, du \]

can be evaluated using standard techniques (e.g., using the Weierstrass substitution or recognizing it as a standard integral).

The result of this integral is:

\[ \int \frac{1}{2 - \cos u} \, du = \frac{u}{\sqrt{3}} + C. \]

Evaluating from \(-\frac{\pi}{2}\) to \(\frac{\pi}{2}\):

\[ \left[ \frac{u}{\sqrt{3}} \right]_{-\frac{\pi}{2}}^{\frac{\pi}{2}} = \frac{\frac{\pi}{2}}{\sqrt{3}} - \left(-\frac{\frac{\pi}{2}}{\sqrt{3}}\right) = \frac{\pi}{\sqrt{3}}. \]

Thus,

\[ \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{1}{2 - \cos 2x} \, dx = \frac{1}{2} \cdot \frac{\pi}{\sqrt{3}} = \frac{\pi}{2\sqrt{3}}. \]

### Step 5: Final calculation

Substituting back into our expression for \(2I\):

\[ 2I = \frac{\pi}{2} \cdot \frac{\pi}{2\sqrt{3}} = \frac{\pi^2}{4\sqrt{3}}. \]

Thus,

\[ I = \frac{\pi^2}{8\sqrt{3}}. \]

### Conclusion

The value of the integral is

\[ \frac{\pi^2}{8\sqrt{3}}. \]