Can someone explain why this is true?
$\int_{0}^{2\pi }\left(\sin nx-\sin kx\right)^2dx=2\pi$ for natural numbers $n,k$ with $n
Can someone explain why this is true?
$\int_{0}^{2\pi }\left(\sin nx-\sin kx\right)^2dx=2\pi$ for natural numbers $n,k$ with $n
What you need here is the following fact:
For positive integers $a$ and $b$ , we have $ \int_0^{2 \pi} \sin (ax) \cdot \sin (bx) \, dx = \begin{cases} 0, &a \neq b, \\ \pi, &a = b. \end{cases} $
Proof. Suppose $a \neq b$; then both $a-b$ and $a+b$ are nonzero. $ \begin{align*} \int_0^{2 \pi} \sin (ax) \cdot \sin (bx) \, dx &= \int_0^{2 \pi} \frac{\cos((a-b)x) - \cos ((a+b)x) }{2} \, dx \\ &= \frac{1}{2} \left[ \frac{\sin ((a-b)x)}{a-b} - \frac{\sin ((a+b)x)}{a+b} \right]_{x=0}^{x = 2 \pi} \\ &= 0, \end{align*} $ since $\sin k\pi = 0$. On the other hand, if $b=a$, then $ \begin{align*} \int_0^{2 \pi} \sin^2 (ax) \, dx &= \int_0^{2 \pi} \frac{1 - \cos (2ax) }{2} \, dx \\ &= \frac{1}{2} \left[ x - \frac{\sin (2ax)}{2a} \right]_{x=0}^{x = 2 \pi} \\ &= \pi. \end{align*} $
Having established this fact, all you need to do is to expand out the product inside the integral and integrate term by term: $ \int_0^{2 \pi} \left( \sin^2 nx + \sin^2 kx - 2 \sin nx \cdot \sin kx \right) \, dx = \pi + \pi - 2 \cdot 0 = 2 \pi. $