## Thomas' Calculus 13th Edition

$\dfrac{2}{k}$
Since, the function $\sin x$ is a periodical function having period $\dfrac{2\pi}{k}$. Thus, we have $A=\int_0^{\pi/k} \sin kx dx$ $\implies [(\dfrac{-1}{k}) \cos kx]_0^{\pi/k}=\dfrac{-1}{k}\cos k(\dfrac{1}{k})-(\dfrac{-1}{k})(\cos 0)$ $\implies A=\dfrac{2}{k}$