Question

If \(m,n\in \mathbb{Z}\), then \[ \int_0^{2\pi}\cos mx\cos nx\,dx+\int_{-\pi}^{\pi}\sin mx\cos nx\,dx= \]

• A. \(0,\ \text{if }\; m=n,\; m,n\in\mathbb{Z}\)
• B. \(\pi,\ \text{if } \; m=n,\; m,n\in\mathbb{Z}\)
• C. \(\pi,\ \text{if }\; m\neq n,\; m,n\in\mathbb{Z}\)
• D. \(2\pi,\ \forall\, m,n\in\mathbb{R}\)

Hint:

Use orthogonality of trigonometric functions: \[ \int_0^{2\pi}\cos mx\cos nx\,dx=0 \text{ if } m\neq n \] and \[ \int_0^{2\pi}\cos^2 mx\,dx=\pi. \] Also, the integral of an odd function over symmetric limits is zero.

Answer 1

The Correct Option is B

Step 1: Consider the first integral.
\[ I_1=\int_0^{2\pi}\cos mx\cos nx\,dx \] Using the orthogonality property, \[ \int_0^{2\pi}\cos mx\cos nx\,dx=0,\quad m\neq n \] and \[ \int_0^{2\pi}\cos^2 mx\,dx=\pi,\quad m=n \] Therefore, \[ I_1=\pi,\quad m=n \]

Step 2: Consider the second integral.
\[ I_2=\int_{-\pi}^{\pi}\sin mx\cos nx\,dx \] Here, \[ \sin mx \] is an odd function and \[ \cos nx \] is an even function.
So, \[ \sin mx\cos nx \] is an odd function.
The integral of an odd function over \([-\pi,\pi]\) is zero.
Hence, \[ I_2=0 \]

Step 3: Add both integrals.
\[ I_1+I_2=\pi+0 \] \[ I_1+I_2=\pi \]

Step 4: Final conclusion.
Therefore, \[ \boxed{\pi,\ \text{if }m=n,\ m,n\in\mathbb{Z}} \]