`int"tan"^(2)(x)/(2)dx=?`

Options

A. `"tan"(x)/(2)-x+C`
B. `"tan"(x)/(2)+x+C`
C. `2"tan"(x)/(2)+x+C`
D. `2"tan"(x)/(2)-x+C`

Explanation

To solve the integral \(\int \frac{\tan^2(x)}{2} \, dx\), we can follow these steps:

### Step 1: Rewrite the Integral
We start by rewriting the integral using the identity for \(\tan^2(x)\):
\[ \tan^2(x) = \sec^2(x) - 1 \]
Thus, we can express the integral as:
\[ \int \frac{\tan^2(x)}{2} \, dx = \int \frac{\sec^2(x) - 1}{2} \, dx \]
This can be separated into two integrals:
\[ = \frac{1}{2} \int \sec^2(x) \, dx - \frac{1}{2} \int 1 \, dx \]

### Step 2: Integrate Each Part
Now, we can integrate each part separately:
1. The integral of \(\sec^2(x)\) is \(\tan(x)\):
\[ \int \sec^2(x) \, dx = \tan(x) \]
2. The integral of \(1\) is simply \(x\):
\[ \int 1 \, dx = x \]

Putting these results back into our expression, we have:
\[ \frac{1}{2} \tan(x) - \frac{1}{2} x + C \]

### Final Answer
Thus, the final result of the integral is:
\[ \int \frac{\tan^2(x)}{2} \, dx = \frac{1}{2} \tan(x) - \frac{1}{2} x + C \]