Question

The value of the integral${\int }_0^{\mathrm{\pi }}|\sin 2x|dx$ is

Answer 1

Find the value of the given integral.

Given integral: ${\int }_0^{\mathrm{\pi }}|\sin 2x|dx$

Let, $2x=z$

So, $2dx=dz$

$\Rightarrow dx=\frac{dz}{2}$

$$\begin{gathered} \therefore {\int }_0^{\mathrm{\pi }}\left|\sin 2x\right|dx=\frac{1}{2}{\int }_0^{2\mathrm{\pi }}\left|\sin z\right|dz \\ =\frac{1}{2}{\int }_0^{4\times \frac{\mathrm{\pi }}{2}}\left|\sin z\right|dz \\ =\frac{4}{2}{\int }_0^{\frac{\mathrm{\pi }}{2}}\sin zdz\left[\because {\int }_0^{\mathrm{nT}}\left|f\left(t\right)\right|dt=n{\int }_0^{T}f\left(t\right)dt,\mathrm{for}\left|\sin z\right|,T=\frac{\mathrm{\pi }}{2}\right] \\ =2{\int }_0^{\frac{\mathrm{\pi }}{2}}\sin zdz \\ =2{\left|-\cos z\right|}_0^{\frac{\mathrm{\pi }}{2}} \\ =2\left[-\cos \frac{\mathrm{\pi }}{2}-\left(-\cos 0\right)\right] \\ =2\left[-0-\left(-1\right)\right] \\ =2 \end{gathered}$$

Hence, the value of the integral${\int }_0^{\mathrm{\pi }}|\sin 2x|dx$ is $2$.