`int_(-pi)^(pi) [cos px-sin qx]^(2)` dx where p,q are integers is equal to

Options

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

Answer

To solve the integral \( I = \int_{-\pi}^{\pi} [\cos(px) - \sin(qx)]^2 \, dx \), where \( p \) and \( q \) are integers, we can follow these steps:

### Step 1: Expand the integrand
We start by expanding the expression inside the integral:
\[ I = \int_{-\pi}^{\pi} [\cos(px) - \sin(qx)]^2 \, dx \]
Using the identity \( (a - b)^2 = a^2 - 2ab + b^2 \), we can rewrite the integrand:
\[ I = \int_{-\pi}^{\pi} [\cos^2(px) - 2\cos(px)\sin(qx) + \sin^2(qx)] \, dx \]

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

### Step 3: Evaluate the integral of the odd function
The term \( \int_{-\pi}^{\pi} \cos(px)\sin(qx) \, dx \) is zero because the integrand is an odd function (the product of an even function \( \cos(px) \) and an odd function \( \sin(qx) \)):
\[ \int_{-\pi}^{\pi} \cos(px)\sin(qx) \, dx = 0 \]

### Step 4: Evaluate the remaining integrals
Now we need to evaluate the two remaining integrals:
1. \( \int_{-\pi}^{\pi} \cos^2(px) \, dx \)
2. \( \int_{-\pi}^{\pi} \sin^2(qx) \, dx \)

Using the identities:
\[ \cos^2(px) = \frac{1 + \cos(2px)}{2}, \quad \sin^2(qx) = \frac{1 - \cos(2qx)}{2} \]

We can rewrite the integrals:
\[ \int_{-\pi}^{\pi} \cos^2(px) \, dx = \int_{-\pi}^{\pi} \frac{1 + \cos(2px)}{2} \, dx = \frac{1}{2} \left( \int_{-\pi}^{\pi} 1 \, dx + \int_{-\pi}^{\pi} \cos(2px) \, dx \right) \]
The integral of \( 1 \) over the interval is:
\[ \int_{-\pi}^{\pi} 1 \, dx = 2\pi \]
And since \( \cos(2px) \) is also an even function, its integral over a symmetric interval around zero is zero:
\[ \int_{-\pi}^{\pi} \cos(2px) \, dx = 0 \]
Thus, we have:
\[ \int_{-\pi}^{\pi} \cos^2(px) \, dx = \frac{1}{2} (2\pi + 0) = \pi \]

Similarly, for \( \sin^2(qx) \):
\[ \int_{-\pi}^{\pi} \sin^2(qx) \, dx = \int_{-\pi}^{\pi} \frac{1 - \cos(2qx)}{2} \, dx = \frac{1}{2} \left( \int_{-\pi}^{\pi} 1 \, dx - \int_{-\pi}^{\pi} \cos(2qx) \, dx \right) \]
Again, we find:
\[ \int_{-\pi}^{\pi} \sin^2(qx) \, dx = \frac{1}{2} (2\pi - 0) = \pi \]

### Step 5: Combine the results
Now we can combine the results:
\[ I = \pi - 0 + \pi = 2\pi \]

### Final Result
Thus, the value of the integral is:
\[ \int_{-\pi}^{\pi} [\cos(px) - \sin(qx)]^2 \, dx = 2\pi \]