IIT JEE , AIEEE Forum - Mathematics , Integral Calculus

Greatdreams

$For\;these \;sort\;of\;problems\;you\;have\;to\;take :$

$x = a{cos}^{2}t + b{sin}^{2}t$

$Hence\;x-a\;=\;b{sin}^{2}t - a(1-{cos}^{2}t)$

$= (b - a){sin}^{2}t$

$Similarly ;\;b-x\;=(b-a){cos}^{2}t$

$Hence \;dx =\;2(b-a)sint\;cost\;dt$

$x = a ; \;sint = 0 \;and\; x = b ;cost=0$

$Thus I = \int_{0}^{\pi/2} 2(b-a)sint costdt / (b-a)sintcost$

$= 2\;[ t ]^{\pi/2}_0$

$= \pi$

$Aliter;\; take\;x-a\;=t^{2}$

$adjust\; limits\;and\;thus; b-x = b-a-t^{2}$

$Hence; I =\int_{0}^{\sqrt{b-a}} 2tdt/(t)\sqrt{b-a-t^{2}}$

$= 2 {sin}^{-1}[ t/\sqrt{b-a}]$

$= 2[ sin^{-1}1 -sin^{-1}0]$

$= 2.\pi/2$

$= \pi$