If `int_(0)^(oo)(sinx)/(x)dx=k`, then the value of `int_(0)^(oo)(sin^(3)x)/(x)dx` is equal to

Options

A. k
B. `(k)/(2)`
C. `(k)/(4)`
D. 2k

Answer

To solve the problem, we need to evaluate the integral \( \int_0^{\infty} \frac{\sin^3 x}{x} \, dx \) given that \( \int_0^{\infty} \frac{\sin x}{x} \, dx = k \).

### Step-by-Step Solution:

1. **Use the identity for \( \sin^3 x \)**:
We can express \( \sin^3 x \) in terms of \( \sin x \) and \( \sin 3x \) using the identity:
\[ \sin^3 x = \frac{3 \sin x - \sin 3x}{4} \]

2. **Substitute the identity into the integral**:
Substitute this identity into the integral:
\[ \int_0^{\infty} \frac{\sin^3 x}{x} \, dx = \int_0^{\infty} \frac{3 \sin x - \sin 3x}{4x} \, dx \]

3. **Split the integral**:
This can be split into two separate integrals:
\[ \int_0^{\infty} \frac{\sin^3 x}{x} \, dx = \frac{1}{4} \left( 3 \int_0^{\infty} \frac{\sin x}{x} \, dx - \int_0^{\infty} \frac{\sin 3x}{x} \, dx \right) \]

4. **Evaluate the integrals**:
We know from the problem statement that:
\[ \int_0^{\infty} \frac{\sin x}{x} \, dx = k \]
For the second integral, we can use a substitution \( t = 3x \), which gives \( dt = 3dx \) or \( dx = \frac{dt}{3} \). The limits remain the same (from 0 to \( \infty \)):
\[ \int_0^{\infty} \frac{\sin 3x}{x} \, dx = \int_0^{\infty} \frac{\sin t}{t} \cdot \frac{1}{3} \, dt = \frac{1}{3} \int_0^{\infty} \frac{\sin t}{t} \, dt = \frac{k}{3} \]

5. **Substitute back into the expression**:
Now substituting back into our expression:
\[ \int_0^{\infty} \frac{\sin^3 x}{x} \, dx = \frac{1}{4} \left( 3k - \frac{k}{3} \right) \]

6. **Simplify the expression**:
To simplify \( 3k - \frac{k}{3} \):
\[ 3k - \frac{k}{3} = \frac{9k}{3} - \frac{k}{3} = \frac{8k}{3} \]
Thus,
\[ \int_0^{\infty} \frac{\sin^3 x}{x} \, dx = \frac{1}{4} \cdot \frac{8k}{3} = \frac{2k}{3} \]

### Final Result:
The value of \( \int_0^{\infty} \frac{\sin^3 x}{x} \, dx \) is \( \frac{2k}{3} \).