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 \( I = \int_{0}^{\frac{\pi}{2}} \sin^3 x \, dx \), we can use the identity for \( \sin^3 x \).

### Step 1: Use the identity for \( \sin^3 x \)
We know that:
\[ \sin^3 x = \frac{3\sin x - \sin 3x}{4} \]
Thus, we can rewrite the integral as:
\[ I = \int_{0}^{\frac{\pi}{2}} \sin^3 x \, dx = \int_{0}^{\frac{\pi}{2}} \frac{3\sin x - \sin 3x}{4} \, dx \]

### Step 2: Split the integral
Now, we can split the integral:
\[ I = \frac{1}{4} \int_{0}^{\frac{\pi}{2}} (3\sin x - \sin 3x) \, dx = \frac{3}{4} \int_{0}^{\frac{\pi}{2}} \sin x \, dx - \frac{1}{4} \int_{0}^{\frac{\pi}{2}} \sin 3x \, dx \]

### Step 3: Evaluate the first integral
The integral of \( \sin x \) is:
\[ \int \sin x \, dx = -\cos x \]
Thus,
\[ \int_{0}^{\frac{\pi}{2}} \sin x \, dx = [-\cos x]_{0}^{\frac{\pi}{2}} = -\cos\left(\frac{\pi}{2}\right) - (-\cos(0)) = 0 - (-1) = 1 \]

### Step 4: Evaluate the second integral
Now, for \( \int \sin 3x \, dx \):
\[ \int \sin 3x \, dx = -\frac{1}{3} \cos 3x \]
Thus,
\[ \int_{0}^{\frac{\pi}{2}} \sin 3x \, dx = \left[-\frac{1}{3} \cos 3x\right]_{0}^{\frac{\pi}{2}} = -\frac{1}{3} \left(\cos\left(\frac{3\pi}{2}\right) - \cos(0)\right) = -\frac{1}{3} (0 - 1) = \frac{1}{3} \]

### Step 5: Substitute back into the integral
Now substituting back into our expression for \( I \):
\[ I = \frac{3}{4} \cdot 1 - \frac{1}{4} \cdot \frac{1}{3} = \frac{3}{4} - \frac{1}{12} \]

### Step 6: Find a common denominator and simplify
To combine these fractions, we find a common denominator (which is 12):
\[ I = \frac{3 \cdot 3}{12} - \frac{1}{12} = \frac{9}{12} - \frac{1}{12} = \frac{8}{12} = \frac{2}{3} \]

### Final Result
Thus, the value of the integral is:
\[ \boxed{\frac{2}{3}} \]