# Solve the following integral. ..............................................

venkat
105 Points
4 years ago
In these kind of sums the first thing you have to do is to put x=t2
then tx=2tdt
substitute it in the integral you will get
$\int \sqrt{(1-t)/(1+t)} \ 2tdt$
Now rationalise the denominator with 1-t
$2\int t-t^2/\sqrt{(1-t^2)} \ dt$
now split the integral into two parts
$\int 2tdt/\sqrt{(1-t^2)} \ -2\int{t^{2}dt/\sqrt{1-t^{2}}}$
We know that integral of f ‘(x)/sqrt f(x) =2f(x)^1/2 and adding ,subtracting 1 in the second part of the integral we get
$2\sqrt{(1-t^2)} \ +2\int{(1-t^{2}+1)dt/\sqrt{1-t^{2}}}$
$2\sqrt{(1-t^2)} \ +2\int\sqrt{(1-t^{2})}dt-2\int dt/\sqrt{1-t^{2}}$
Then by using the identites we get that
$2\sqrt{(1-t^2)} + \sqrt{1-t^{2}}-sin^{-1}\sqrt{1-t^{2}}$
now replacing t2 with x we get
$2\sqrt{(1-x)} + \sqrt{1-x}-sin^{-1}\sqrt{1-x}+c$