Question

${\int }_0^{\pi }|\sin 8x|dx$

Answer 1

${\int }_0^{\pi }|\sin 8x|\cdot dx$

Period of $|\sin 8x|$ is $\pi /8$

${\int }_0^{mT}f\left(x\right)\cdot dx=m{\int }_0^{T}f\left(x\right)\cdot dx$

where T is period of $f\left(x\right)$ $\to$ Main logic

$={\int }_o^{8\times \pi /8}|\sin 8x|\cdot dx$

$=8{\int }_o^{\pi /8}|\sin 8x|\cdot dx$

In $0<x<\frac{\pi }{8}$

$0<8x<\pi$

$0<\sin 8x$

$\therefore |\sin 8x|=\sin 8x$

$=8{\int }_o^{\pi /8}\sin 8x\cdot dx$

$=8\times {\left[\frac{-\cos 8x}{8}\right]}_o^{\pi /8}$

$=8[+\frac{1}{8}+\frac{1}{8}]=2$.