Question

Evaluate : ${\int }_0^{2\pi }\left[2\sin x\right]dx$

Answer 1

We have

$\begin{array}{c}I={\int }_0^{2\pi }\left[2\sin xdx\right]\\ =2{\int }_0^{2\pi }sinxdx\\ =2{\left[-\cos x\right]}_0^{2\pi }\\ =-2\cos \left(2\pi \right)-2\cos \left(0\right)\\ =-2-2\left(\cos 2\pi =1\right)\\ =-4\\ Hence,Itistherequiredanswer.\end{array}$