Evaluate `intsin^2xdx`

Answer

To evaluate the integral \(\int \sin^2 x \, dx\), we can use a trigonometric identity to simplify the expression. Here's the step-by-step solution:

### Step 1: Use the Trigonometric Identity
We know that:
\[ \sin^2 x = \frac{1 - \cos 2x}{2} \]
This allows us to rewrite the integral:
\[ \int \sin^2 x \, dx = \int \frac{1 - \cos 2x}{2} \, dx \]

### Step 2: Simplify the Integral
Now, we can factor out the \(\frac{1}{2}\):
\[ \int \sin^2 x \, dx = \frac{1}{2} \int (1 - \cos 2x) \, dx \]

### Step 3: Split the Integral
We can split the integral into two parts:
\[ \int (1 - \cos 2x) \, dx = \int 1 \, dx - \int \cos 2x \, dx \]

### Step 4: Integrate Each Part
Now we will integrate each part separately:
1. The integral of \(1\) is:
\[ \int 1 \, dx = x \]

2. The integral of \(\cos 2x\) requires a substitution. The integral is:
\[ \int \cos 2x \, dx = \frac{1}{2} \sin 2x \]
(Here we divide by the derivative of \(2x\), which is \(2\)).

### Step 5: Combine the Results
Putting it all together, we have:
\[ \int (1 - \cos 2x) \, dx = x - \frac{1}{2} \sin 2x \]

### Step 6: Multiply by \(\frac{1}{2}\)
Now, we go back to our earlier expression:
\[ \int \sin^2 x \, dx = \frac{1}{2} \left( x - \frac{1}{2} \sin 2x \right) \]
This simplifies to:
\[ \int \sin^2 x \, dx = \frac{x}{2} - \frac{1}{4} \sin 2x + C \]

### Final Answer
Thus, the evaluated integral is:
\[ \int \sin^2 x \, dx = \frac{x}{2} - \frac{1}{4} \sin 2x + C \]