The value of the integral ` int_(0) ^(pi//2) sin ^3 x dx ` is :

Options

A. ` (3)/(8) pi `
B. 0
C. ` (3 )/(16) pi `
D. None of these

Answer

To solve the integral \( \int_{0}^{\frac{\pi}{2}} \sin^3 x \, dx \), we can use the reduction formula or a trigonometric identity. Here’s a step-by-step solution:

### Step 1: Use the identity for \( \sin^3 x \)
We can express \( \sin^3 x \) in terms of \( \sin x \) and \( \cos x \):
\[ \sin^3 x = \sin x (1 - \cos^2 x) = \sin x - \sin x \cos^2 x \]

### Step 2: Rewrite the integral
Now we can rewrite the integral:
\[ \int_{0}^{\frac{\pi}{2}} \sin^3 x \, dx = \int_{0}^{\frac{\pi}{2}} \sin x \, dx - \int_{0}^{\frac{\pi}{2}} \sin x \cos^2 x \, dx \]

### Step 3: Evaluate the first integral
The first integral \( \int_{0}^{\frac{\pi}{2}} \sin x \, dx \) is straightforward:
\[ \int \sin x \, dx = -\cos x \quad \text{(evaluating from 0 to } \frac{\pi}{2}\text{)} \]
\[ = -\cos\left(\frac{\pi}{2}\right) + \cos(0) = 0 + 1 = 1 \]

### Step 4: Evaluate the second integral
For the second integral \( \int_{0}^{\frac{\pi}{2}} \sin x \cos^2 x \, dx \), we can use the substitution \( u = \cos x \), which gives \( du = -\sin x \, dx \). Changing the limits accordingly:
- When \( x = 0 \), \( u = 1 \)
- When \( x = \frac{\pi}{2} \), \( u = 0 \)

Thus, the integral becomes:
\[ \int_{0}^{\frac{\pi}{2}} \sin x \cos^2 x \, dx = -\int_{1}^{0} u^2 \, du = \int_{0}^{1} u^2 \, du \]
\[ = \left[ \frac{u^3}{3} \right]_{0}^{1} = \frac{1^3}{3} - 0 = \frac{1}{3} \]

### Step 5: Combine the results
Now, substituting back into our equation:
\[ \int_{0}^{\frac{\pi}{2}} \sin^3 x \, dx = 1 - \frac{1}{3} = \frac{2}{3} \]

### Final Answer
Thus, the value of the integral \( \int_{0}^{\frac{\pi}{2}} \sin^3 x \, dx \) is:
\[ \boxed{\frac{2}{3}} \]