If `int_0^oosinx/xdx=pi/2`, then `int_0^oosin^3x/xdx` is equal to

Options

A. `pi//2`
B. `pi//4`
C. `pi//6`
D. `3pi//2`

Explanation

To solve the problem, we need to find the value of the integral \( \int_0^{\infty} \frac{\sin^3 x}{x} \, dx \) given that \( \int_0^{\infty} \frac{\sin x}{x} \, dx = \frac{\pi}{2} \).

### Step-by-Step Solution:

1. **Given Information**:
We have:
\[ I = \int_0^{\infty} \frac{\sin x}{x} \, dx = \frac{\pi}{2} \]

2. **Express \( \sin^3 x \)**:
We can use the identity for \( \sin^3 x \):
\[ \sin^3 x = \frac{3 \sin x - \sin(3x)}{4} \]
Therefore, we can rewrite the integral \( I' \):
\[ I' = \int_0^{\infty} \frac{\sin^3 x}{x} \, dx = \int_0^{\infty} \frac{3 \sin x - \sin(3x)}{4x} \, dx \]

3. **Separate the Integral**:
We can separate the integral into two parts:
\[ I' = \frac{1}{4} \left( 3 \int_0^{\infty} \frac{\sin x}{x} \, dx - \int_0^{\infty} \frac{\sin(3x)}{x} \, dx \right) \]

4. **Evaluate Each Integral**:
We already know:
\[ \int_0^{\infty} \frac{\sin x}{x} \, dx = I = \frac{\pi}{2} \]
For the second integral, we can use the substitution \( u = 3x \), which gives \( du = 3dx \) or \( dx = \frac{du}{3} \):
\[ \int_0^{\infty} \frac{\sin(3x)}{x} \, dx = \int_0^{\infty} \frac{\sin u}{u} \cdot 3 \, \frac{du}{3} = \int_0^{\infty} \frac{\sin u}{u} \, du = \frac{\pi}{2} \]

5. **Substituting Back**:
Now we substitute back into our expression for \( I' \):
\[ I' = \frac{1}{4} \left( 3 \cdot \frac{\pi}{2} - \frac{\pi}{2} \right) \]
Simplifying this gives:
\[ I' = \frac{1}{4} \left( \frac{3\pi}{2} - \frac{\pi}{2} \right) = \frac{1}{4} \left( \frac{2\pi}{2} \right) = \frac{1}{4} \cdot \pi = \frac{\pi}{4} \]

6. **Final Result**:
Therefore, the value of the integral \( \int_0^{\infty} \frac{\sin^3 x}{x} \, dx \) is:
\[ \int_0^{\infty} \frac{\sin^3 x}{x} \, dx = \frac{\pi}{4} \]