`intsin^(2)x.cos^(2)xdx=`

Options

A. `(x)/(8)+(sin4x)/(32)+c`
B. `-(x)/(8)-(sin4x)/(32)+c`
C. `(sin4x)/(32)-(x)/(8)+c`
D. `(x)/(8)-(sin4x)/(32)+c`

Answer

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

### Step 1: Rewrite the Integral
We start with the integral:
\[ I = \int \sin^2 x \cos^2 x \, dx \]

### Step 2: Use the Trigonometric Identity
We can use the identity \( \sin^2 x = \frac{1 - \cos(2x)}{2} \) and \( \cos^2 x = \frac{1 + \cos(2x)}{2} \) to rewrite the integral:
\[ I = \int \left(\frac{1 - \cos(2x)}{2}\right) \left(\frac{1 + \cos(2x)}{2}\right) \, dx \]

### Step 3: Simplify the Expression
Now, we simplify the expression:
\[ I = \int \frac{(1 - \cos(2x))(1 + \cos(2x))}{4} \, dx \]
\[ = \int \frac{1 - \cos^2(2x)}{4} \, dx \]
Using the identity \( \sin^2(2x) = 1 - \cos^2(2x) \):
\[ I = \int \frac{\sin^2(2x)}{4} \, dx \]

### Step 4: Factor Out the Constant
We can factor out the constant \( \frac{1}{4} \):
\[ I = \frac{1}{4} \int \sin^2(2x) \, dx \]

### Step 5: Use the Identity for \( \sin^2 \)
Using the identity \( \sin^2(2x) = \frac{1 - \cos(4x)}{2} \):
\[ I = \frac{1}{4} \int \frac{1 - \cos(4x)}{2} \, dx \]
\[ = \frac{1}{8} \int (1 - \cos(4x)) \, dx \]

### Step 6: Integrate
Now we can integrate:
\[ I = \frac{1}{8} \left( \int 1 \, dx - \int \cos(4x) \, dx \right) \]
\[ = \frac{1}{8} \left( x - \frac{\sin(4x)}{4} \right) + C \]
\[ = \frac{x}{8} - \frac{\sin(4x)}{32} + C \]

### Final Answer
Thus, the final result of the integral is:
\[ I = \frac{x}{8} - \frac{\sin(4x)}{32} + C \]