Evaluate : `int_((pi)/(6))^((pi)/(3)) sin^(2) x dx`

Answer

To evaluate the integral \( \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \sin^2 x \, dx \), we can follow these steps:

### Step 1: Rewrite the Integral
We start with the integral:
\[ I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \sin^2 x \, dx \]

### Step 2: Use the Identity for \(\sin^2 x\)
We can use the trigonometric identity:
\[ \sin^2 x = \frac{1 - \cos 2x}{2} \]
Substituting this into the integral gives:
\[ I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{1 - \cos 2x}{2} \, dx \]

### Step 3: Simplify the Integral
This can be simplified as:
\[ I = \frac{1}{2} \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} (1 - \cos 2x) \, dx \]
We can separate the integral:
\[ I = \frac{1}{2} \left( \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} 1 \, dx - \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \cos 2x \, dx \right) \]

### Step 4: Evaluate Each Integral
1. The first integral:
\[ \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} 1 \, dx = \left[ x \right]_{\frac{\pi}{6}}^{\frac{\pi}{3}} = \frac{\pi}{3} - \frac{\pi}{6} = \frac{\pi}{6} \]

2. The second integral:
\[ \int \cos 2x \, dx = \frac{1}{2} \sin 2x \]
Therefore,
\[ \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \cos 2x \, dx = \left[ \frac{1}{2} \sin 2x \right]_{\frac{\pi}{6}}^{\frac{\pi}{3}} = \frac{1}{2} \left( \sin \frac{2\pi}{3} - \sin \frac{\pi}{3} \right) \]
We know that:
\[ \sin \frac{2\pi}{3} = \sin \left( \pi - \frac{\pi}{3} \right) = \sin \frac{\pi}{3} = \frac{\sqrt{3}}{2} \]
Thus,
\[ \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \cos 2x \, dx = \frac{1}{2} \left( \frac{\sqrt{3}}{2} - \frac{\sqrt{3}}{2} \right) = 0 \]

### Step 5: Combine Results
Now substituting back into our expression for \(I\):
\[ I = \frac{1}{2} \left( \frac{\pi}{6} - 0 \right) = \frac{\pi}{12} \]

### Final Answer
Thus, the value of the integral is:
\[ \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \sin^2 x \, dx = \frac{\pi}{12} \]

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