Question

Solve:
${\int }_0^{T/2}{I}_0\sin \left(wt\right)dt$

Answer 1

${\int }_0^{T/2}{I}_{0}sin\left(wt\right).dt$

$=I_0{\int }_0^{T/2}sin\left(wt\right).dt$

$=I_0[\frac{-coswt}{w}{]}_0^{T/2}$

$=\frac{I_0}{w}[-cosw.\frac{T}{2}-(-cos0\left)\right]$

$=\frac{I_0}{w}[-cos(\frac{2\pi }{T}.\frac{T}{2})+cos0]$

$=\frac{I_0}{w}[-cos\pi +cos0]$

$=\frac{I_0}{w}[I+1]$

$=\frac{2I_0}{w}$

$=\frac{2I_0}{\left(\frac{2\pi }{T}\right)}$

$=\frac{T.I_0}{\pi }$