`int_(0)^(pi)(cos^(4)x-sin^(4)x)dx=`

Options

A. 2
B. `-2`
C. 0
D. 4

Answer

To solve the integral \( I = \int_{0}^{\pi} (\cos^4 x - \sin^4 x) \, dx \), we can follow these steps:

### Step 1: Rewrite the expression
We can use the identity \( a^2 - b^2 = (a-b)(a+b) \) to rewrite \( \cos^4 x - \sin^4 x \):
\[ \cos^4 x - \sin^4 x = (\cos^2 x - \sin^2 x)(\cos^2 x + \sin^2 x) \]
Since \( \cos^2 x + \sin^2 x = 1 \), we have:
\[ \cos^4 x - \sin^4 x = \cos^2 x - \sin^2 x \]

### Step 2: Substitute into the integral
Now we can substitute this back into the integral:
\[ I = \int_{0}^{\pi} (\cos^2 x - \sin^2 x) \, dx \]

### Step 3: Use the identity for sine
We know that \( \sin^2 x = 1 - \cos^2 x \). Therefore:
\[ \cos^2 x - \sin^2 x = \cos^2 x - (1 - \cos^2 x) = 2\cos^2 x - 1 \]
This gives us:
\[ I = \int_{0}^{\pi} (2\cos^2 x - 1) \, dx \]

### Step 4: Split the integral
We can split the integral into two parts:
\[ I = 2 \int_{0}^{\pi} \cos^2 x \, dx - \int_{0}^{\pi} 1 \, dx \]

### Step 5: Evaluate the integrals
The integral \( \int_{0}^{\pi} 1 \, dx = \pi \).

For \( \int_{0}^{\pi} \cos^2 x \, dx \), we can use the identity:
\[ \cos^2 x = \frac{1 + \cos 2x}{2} \]
Thus,
\[ \int_{0}^{\pi} \cos^2 x \, dx = \int_{0}^{\pi} \frac{1 + \cos 2x}{2} \, dx = \frac{1}{2} \left( \int_{0}^{\pi} 1 \, dx + \int_{0}^{\pi} \cos 2x \, dx \right) \]
The first part is \( \frac{1}{2} \cdot \pi \). The second part, \( \int_{0}^{\pi} \cos 2x \, dx = 0 \) (since the integral of cosine over a full period is zero).

So,
\[ \int_{0}^{\pi} \cos^2 x \, dx = \frac{1}{2} \cdot \pi = \frac{\pi}{2} \]

### Step 6: Substitute back into the equation for \( I \)
Now substituting back:
\[ I = 2 \cdot \frac{\pi}{2} - \pi = \pi - \pi = 0 \]

### Final Answer
Thus, the value of the integral is:
\[ \boxed{0} \]