`intsin^(2)4xdx=`

Options

A. `(x)/(2)+(sin8x)/(16)+c`
B. `(x)/(2)-(sin8x)/(16)-c`
C. `cos^(2)4x+c`
D. `4sin8x+c`

Answer

To solve the integral \(\int \sin^2(4x) \, dx\), we can follow these steps:

### Step 1: Use the identity for \(\sin^2\)
We can use the trigonometric identity:
\[ \sin^2(\theta) = \frac{1 - \cos(2\theta)}{2} \]
In our case, \(\theta = 4x\), so:
\[ \sin^2(4x) = \frac{1 - \cos(8x)}{2} \]

### Step 2: Substitute into the integral
Now we can rewrite the integral:
\[ \int \sin^2(4x) \, dx = \int \frac{1 - \cos(8x)}{2} \, dx \]
This simplifies to:
\[ \frac{1}{2} \int (1 - \cos(8x)) \, dx \]

### Step 3: Split the integral
We can separate the integral into two parts:
\[ \frac{1}{2} \left( \int 1 \, dx - \int \cos(8x) \, dx \right) \]

### Step 4: Integrate each part
1. The integral of \(1\) is simply \(x\):
\[ \int 1 \, dx = x \]
2. The integral of \(\cos(8x)\) requires a substitution. The integral is:
\[ \int \cos(8x) \, dx = \frac{\sin(8x)}{8} \]

### Step 5: Combine the results
Now we can combine the results from the two integrals:
\[ \frac{1}{2} \left( x - \frac{\sin(8x)}{8} \right) + C \]
This simplifies to:
\[ \frac{x}{2} - \frac{\sin(8x)}{16} + C \]

### Final Answer
Thus, the final result of the integral is:
\[ \int \sin^2(4x) \, dx = \frac{x}{2} - \frac{\sin(8x)}{16} + C \]