`int_0^(pi//2) sin^3 x dx`

Answer

To solve the integral \( I = \int_0^{\frac{\pi}{2}} \sin^3 x \, dx \), we can use a trigonometric identity to simplify the integration process.

### 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 into two parts:
\[ I = \frac{1}{4} \int_0^{\frac{\pi}{2}} 3 \sin x \, dx - \frac{1}{4} \int_0^{\frac{\pi}{2}} \sin(3x) \, dx \]

### Step 3: Calculate the first integral

The first integral is:
\[ \int_0^{\frac{\pi}{2}} 3 \sin x \, dx = 3 \left[-\cos x\right]_0^{\frac{\pi}{2}} = 3 \left[-\cos\left(\frac{\pi}{2}\right) + \cos(0)\right] = 3 \left[0 + 1\right] = 3 \]

### Step 4: Calculate the second integral

The second integral is:
\[ \int_0^{\frac{\pi}{2}} \sin(3x) \, dx = \left[-\frac{1}{3} \cos(3x)\right]_0^{\frac{\pi}{2}} = -\frac{1}{3} \left[\cos\left(\frac{3\pi}{2}\right) - \cos(0)\right] = -\frac{1}{3} \left[0 - 1\right] = \frac{1}{3} \]

### Step 5: Substitute back into the integral

Now substituting back into the expression for \( I \):
\[ I = \frac{1}{4} \left(3 - \frac{1}{3}\right) = \frac{1}{4} \left(\frac{9}{3} - \frac{1}{3}\right) = \frac{1}{4} \left(\frac{8}{3}\right) = \frac{2}{3} \]

### Final Answer

Thus, the value of the integral is:
\[ I = \frac{2}{3} \]