Question

Show that ${\int }_0^{n\pi +v}|\sin x|dx=2n+1-\cos v$ , where $n$ is a positive interger and $0\le v<\pi$

Answer 1

${\int }_0^{n\pi +v}|sinx|dx={\int }_0^v|sinx|dx+{\int }_v^{n\pi +v}|sinx|dx$

In,${\int }_0^v|sinx|dx$ here $sinx$ is positve as $0\le v<\pi$

${\int }_0^v|sinx|dx={\int }_0^{v}sinxdx=[-cosx{]}_0^v$

$\Rightarrow {\int }_0^v|sinx|dx=-cosv+1$

${\int }_v^{n\pi +v}|sinx|dx=n{\int }_0^{\pi }|sinx|dx$

as ${\int }_a^{a+nT}f\left(x\right)dx=n{\int }_0^{T}f\left(x\right)dx$

${\int }_v^{n\pi +v}|sinx|dx=n{\int }_0^{\pi }sinxdx$

$\therefore =n[-cosx{]}_0^{\pi }=n[-cos\pi -(-cosO)]$

i.e $=n[1+1]=2n$

$\therefore$ RHS $={\int }_0^{n\pi +v}|sinx|dx=-cosv+1+2n$

Thus RHS=LHS. Hence proved