The current in a coil of self inductance L = 2 . 0H is increasing...

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

$∴ W = {∫}_{0}^{t}Lidi$ $= {∫}_{0}^{t}L2sin{t}^{2}d\left(2sin{t}^{2}\right)$

$⇒ W = {∫}_{0}^{t}8Lsin{t}^{2}cos{t}^{2}tdt = 4L{∫}_{0}^{t}\sin 2{t}^{2}dt$

Let $\theta = 2{t}^{2} ⇒ d\theta = 4tdt$
$∴$ the integral
$= 4L ∫ sin\thetadt / 4L ( - cos\theta ) = - Lcos2{t}^{2}$
$∴ W = - {\left[\cos 2{t}^{2}\right]}_{0}^{\sqrt{\pi / 2}} = 2L = 4J$