STD 12 - 7.2 definite integral — Mathematics STD 12 Science — Question

$ = 4\int_0^{\pi /4} {\frac{{2\tan \theta \,{{\sec }^2}\theta \,d\theta }}{{{{\tan }^4}\theta + 1}}} $

{Dividing numerator and denominator by ${\cos ^4}\theta $ }

Now put ${\tan ^2}\theta = t $

$\Rightarrow 2\tan \theta \,{\sec ^2}\theta \,d\theta = dt$,  then the reduced form is 

$4\int_0^1 {\frac{{dt}}{{{t^2} + 1}}} = 4[{\tan ^{ - 1}}t]_0^1 = 4\left[ {\frac{1}{4}\pi - 0} \right] = \pi $.