The value of the definite integral `I=int_(-1)^(1)ln((2-sin^(3)x)/(2+sin^(3)x))dx` is equal to

Options

A. ln 4
B. ln 2
C. 0
D. `ln((1)/(2))`

Answer

To solve the definite integral

\[ I = \int_{-1}^{1} \ln\left(\frac{2 - \sin^3 x}{2 + \sin^3 x}\right) dx, \]

we can use the property of definite integrals that states:

\[ \int_{-a}^{a} f(x) \, dx = \int_{-a}^{a} f(-x) \, dx. \]

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

We will first compute \( I \) by substituting \( x \) with \( -x \):

\[ I = \int_{-1}^{1} \ln\left(\frac{2 - \sin^3(-x)}{2 + \sin^3(-x)}\right) dx. \]

Since \( \sin(-x) = -\sin(x) \), we have:

\[ \sin^3(-x) = (-\sin x)^3 = -\sin^3 x. \]

Thus, we can rewrite the integral as:

\[ I = \int_{-1}^{1} \ln\left(\frac{2 + \sin^3 x}{2 - \sin^3 x}\right) dx. \]

### Step 2: Add the two expressions for \( I \)

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

1. \( I = \int_{-1}^{1} \ln\left(\frac{2 - \sin^3 x}{2 + \sin^3 x}\right) dx \)
2. \( I = \int_{-1}^{1} \ln\left(\frac{2 + \sin^3 x}{2 - \sin^3 x}\right) dx \)

Adding these two equations, we get:

\[ 2I = \int_{-1}^{1} \left[ \ln\left(\frac{2 - \sin^3 x}{2 + \sin^3 x}\right) + \ln\left(\frac{2 + \sin^3 x}{2 - \sin^3 x}\right) \right] dx. \]

### Step 3: Simplify the logarithmic expression

Using the property of logarithms \( \ln a + \ln b = \ln(ab) \), we can combine the logarithms:

\[ 2I = \int_{-1}^{1} \ln\left(\frac{(2 - \sin^3 x)(2 + \sin^3 x)}{(2 + \sin^3 x)(2 - \sin^3 x)}\right) dx. \]

This simplifies to:

\[ 2I = \int_{-1}^{1} \ln(1) \, dx. \]

Since \( \ln(1) = 0 \), we have:

\[ 2I = \int_{-1}^{1} 0 \, dx = 0. \]

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

Thus, we find:

\[ I = 0. \]

### Final Answer

The value of the definite integral is:

\[ \boxed{0}. \]