The value of the integral `int_(-pi//2)^(pi//2) sin^(4) x (1+ log ((2+ sinx)/( 2- sin x))) dx` is

Options

A. `(3)/(8) pi`
B. `0`
C. `(3)/( 16) pi`
D. `(3)/(8)`

Answer

To solve the integral

\[ I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin^4 x \left(1 + \log \left(\frac{2 + \sin x}{2 - \sin x}\right)\right) dx, \]

we can break it down into two parts:

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

### Step 1: Evaluate the first integral

The first integral is

\[ I_1 = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin^4 x \, dx. \]

Since \(\sin^4 x\) is an even function, we can simplify this integral:

\[ I_1 = 2 \int_{0}^{\frac{\pi}{2}} \sin^4 x \, dx. \]

### Step 2: Use the power-reduction formula

We can use the power-reduction formula for \(\sin^4 x\):

\[ \sin^4 x = \left(\sin^2 x\right)^2 = \left(\frac{1 - \cos(2x)}{2}\right)^2 = \frac{1 - 2\cos(2x) + \cos^2(2x)}{4}. \]

Using \(\cos^2(2x) = \frac{1 + \cos(4x)}{2}\), we get:

\[ \sin^4 x = \frac{1}{4} \left(1 - 2\cos(2x) + \frac{1 + \cos(4x)}{2}\right) = \frac{1}{8} + \frac{1}{4}\cos(4x) - \frac{1}{2}\cos(2x). \]

### Step 3: Integrate \(I_1\)

Now we can integrate:

\[ I_1 = 2 \int_0^{\frac{\pi}{2}} \left(\frac{1}{8} - \frac{1}{2}\cos(2x) + \frac{1}{8}\cos(4x)\right) dx. \]

Calculating each term separately:

1. \(\int_0^{\frac{\pi}{2}} 1 \, dx = \frac{\pi}{2}\)
2. \(\int_0^{\frac{\pi}{2}} \cos(2x) \, dx = 0\) (since \(\sin(2x)\) evaluated from \(0\) to \(\frac{\pi}{2}\) is \(0\))
3. \(\int_0^{\frac{\pi}{2}} \cos(4x) \, dx = 0\) (similarly, \(\sin(4x)\) evaluated from \(0\) to \(\frac{\pi}{2}\) is \(0\))

Thus,

\[ I_1 = 2 \left(\frac{1}{8} \cdot \frac{\pi}{2} - 0 + 0\right) = \frac{\pi}{8}. \]

### Step 4: Evaluate the second integral

Now we need to evaluate

\[ I_2 = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin^4 x \log \left(\frac{2 + \sin x}{2 - \sin x}\right) dx. \]

Using the property of logarithms, we can rewrite:

\[ \log \left(\frac{2 + \sin x}{2 - \sin x}\right) = \log(2 + \sin x) - \log(2 - \sin x). \]

Now, we check the behavior of \(\log \left(\frac{2 + \sin x}{2 - \sin x}\right)\):

When we replace \(x\) with \(-x\):

\[ \log \left(\frac{2 + \sin(-x)}{2 - \sin(-x)}\right) = \log \left(\frac{2 - \sin x}{2 + \sin x}\right) = -\log \left(\frac{2 + \sin x}{2 - \sin x}\right). \]

This shows that \(\sin^4 x \log \left(\frac{2 + \sin x}{2 - \sin x}\right)\) is an odd function. Therefore,

\[ I_2 = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin^4 x \log \left(\frac{2 + \sin x}{2 - \sin x}\right) dx = 0. \]

### Step 5: Combine results

Finally, we combine both integrals:

\[ I = I_1 + I_2 = \frac{\pi}{8} + 0 = \frac{\pi}{8}. \]

### Final Answer

Thus, the value of the integral is

\[ \boxed{\frac{\pi}{8}}. \]