The current in a coil of self inductance L = 2 . 0   H is increasing according to the law i = 2 sin t 2…

Let the current be $2 A$ at $t=\tau$
Then $2=2sin{\tau }^2\Rightarrow \tau =\sqrt{\pi /2}$
when the instantaneous current is $i$, the
self-induced emf is $L\frac{di}{dt}.$ If $dq$ amount of
charge is displaced in time $dt$, then elementary work done
$=\left(L\frac{di}{dt}\right)dq=L\frac{di}{dt}idt=$$Lidi$

$\therefore W={\int }_0^t Lidi$  $={\int }_0^{t}L2sin{t}^{2}d\left(2sin{t}^2\right)$

$\Rightarrow W={\int }_0^{t}8Lsin{t}^{2}cos{t}^{2}tdt=4L{\int }_0^{t}sin2t^{2}dt$

Let $\theta =2t^2\Rightarrow d\theta =4tdt$
$\therefore$ the integral
$=4L\int sin\theta dt/4L(-cos\theta )=-Lcos2t^2$
$\therefore W=-{\left[cos2t^2\right]}_0^{\sqrt{\pi /2}}=2L=4J$