`int_(-pi)^(pi)sin^(2)x.cos^(2)x dx=`

Options

A. 0
B. `int_(0)^(pi//2)sin^(2)x.cos^(2)x dx`
C. `4 int_(0)^(pi//2)sin^(2)x.cos^(2)x dx`
D. 1

Answer

To solve the integral \( \int_{-\pi}^{\pi} \sin^2 x \cos^2 x \, dx \), we can follow these steps:

### Step 1: Define the function
Let \( f(x) = \sin^2 x \cos^2 x \).

### Step 2: Check for even function
We check if \( f(-x) = f(x) \):
\[ f(-x) = \sin^2(-x) \cos^2(-x) = \sin^2 x \cos^2 x = f(x) \]
Since \( f(-x) = f(x) \), \( f(x) \) is an even function.

### Step 3: Use the property of even functions
For even functions, we can simplify the integral:
\[ \int_{-a}^{a} f(x) \, dx = 2 \int_{0}^{a} f(x) \, dx \]
Thus,
\[ \int_{-\pi}^{\pi} \sin^2 x \cos^2 x \, dx = 2 \int_{0}^{\pi} \sin^2 x \cos^2 x \, dx \]

### Step 4: Simplify the integral
We can further simplify \( \sin^2 x \cos^2 x \):
\[ \sin^2 x \cos^2 x = \frac{1}{4} \sin^2(2x) \]
So,
\[ \int_{0}^{\pi} \sin^2 x \cos^2 x \, dx = \frac{1}{4} \int_{0}^{\pi} \sin^2(2x) \, dx \]

### Step 5: Use the identity for \( \sin^2 \)
Using the identity \( \sin^2 \theta = \frac{1 - \cos(2\theta)}{2} \):
\[ \int_{0}^{\pi} \sin^2(2x) \, dx = \int_{0}^{\pi} \frac{1 - \cos(4x)}{2} \, dx \]
This can be split into two integrals:
\[ = \frac{1}{2} \int_{0}^{\pi} 1 \, dx - \frac{1}{2} \int_{0}^{\pi} \cos(4x) \, dx \]

### Step 6: Evaluate the integrals
The first integral is straightforward:
\[ \int_{0}^{\pi} 1 \, dx = \pi \]
The second integral evaluates to zero, since the integral of \( \cos(4x) \) over a complete period is zero:
\[ \int_{0}^{\pi} \cos(4x) \, dx = 0 \]

### Step 7: Combine results
Thus,
\[ \int_{0}^{\pi} \sin^2(2x) \, dx = \frac{1}{2} \cdot \pi - 0 = \frac{\pi}{2} \]

### Step 8: Substitute back into the integral
Now we substitute back:
\[ \int_{0}^{\pi} \sin^2 x \cos^2 x \, dx = \frac{1}{4} \cdot \frac{\pi}{2} = \frac{\pi}{8} \]

### Step 9: Final result
Finally, we have:
\[ \int_{-\pi}^{\pi} \sin^2 x \cos^2 x \, dx = 2 \cdot \frac{\pi}{8} = \frac{\pi}{4} \]

Thus, the final answer is:
\[ \int_{-\pi}^{\pi} \sin^2 x \cos^2 x \, dx = \frac{\pi}{4} \]

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