The value of integral `int_(-pi)^(pi) (cos ax-sin b x)^(2)dx`, where (a and b are integers), is

Options

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

Answer

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

### Step 1: Expand the integrand
First, we expand the expression inside the integral:
\[ I = \int_{-\pi}^{\pi} (\cos(ax) - \sin(bx))^2 \, dx = \int_{-\pi}^{\pi} (\cos^2(ax) - 2\cos(ax)\sin(bx) + \sin^2(bx)) \, dx \]

### Step 2: Separate the integral
We can separate the integral into three parts:
\[ I = \int_{-\pi}^{\pi} \cos^2(ax) \, dx - 2 \int_{-\pi}^{\pi} \cos(ax)\sin(bx) \, dx + \int_{-\pi}^{\pi} \sin^2(bx) \, dx \]

### Step 3: Evaluate each integral
1. **Evaluate \( \int_{-\pi}^{\pi} \cos^2(ax) \, dx \)**:
Using the identity \( \cos^2(\theta) = \frac{1 + \cos(2\theta)}{2} \):
\[ \int_{-\pi}^{\pi} \cos^2(ax) \, dx = \int_{-\pi}^{\pi} \frac{1 + \cos(2ax)}{2} \, dx = \frac{1}{2} \left( \int_{-\pi}^{\pi} 1 \, dx + \int_{-\pi}^{\pi} \cos(2ax) \, dx \right) \]
The first integral evaluates to \( 2\pi \) and the second integral evaluates to \( 0 \) (since \( \cos(2ax) \) is an even function over a symmetric interval):
\[ \int_{-\pi}^{\pi} \cos^2(ax) \, dx = \frac{1}{2} (2\pi + 0) = \pi \]

2. **Evaluate \( -2 \int_{-\pi}^{\pi} \cos(ax)\sin(bx) \, dx \)**:
The integral \( \int_{-\pi}^{\pi} \cos(ax)\sin(bx) \, dx \) evaluates to \( 0 \) (since \( \cos(ax)\sin(bx) \) is an odd function):
\[ -2 \int_{-\pi}^{\pi} \cos(ax)\sin(bx) \, dx = -2 \cdot 0 = 0 \]

3. **Evaluate \( \int_{-\pi}^{\pi} \sin^2(bx) \, dx \)**:
Using the identity \( \sin^2(\theta) = \frac{1 - \cos(2\theta)}{2} \):
\[ \int_{-\pi}^{\pi} \sin^2(bx) \, dx = \int_{-\pi}^{\pi} \frac{1 - \cos(2bx)}{2} \, dx = \frac{1}{2} \left( \int_{-\pi}^{\pi} 1 \, dx - \int_{-\pi}^{\pi} \cos(2bx) \, dx \right) \]
Similar to before, the first integral evaluates to \( 2\pi \) and the second integral evaluates to \( 0 \):
\[ \int_{-\pi}^{\pi} \sin^2(bx) \, dx = \frac{1}{2} (2\pi - 0) = \pi \]

### Step 4: Combine the results
Now we combine the results from each part:
\[ I = \pi + 0 + \pi = 2\pi \]

### Final Answer
Thus, the value of the integral is:
\[ \boxed{2\pi} \]