Answer to Question #113565 in Calculus for Sourav mondal

Question

Integration of [(a√(a-x))/√(a+x)]

Answer 1

$I=\int{\frac{a\sqrt{a-x}}{\sqrt{a+x}}}dx=a\int{\sqrt{\frac{a-x}{a+x}}}dx$

Let

$\frac{a-x}{a+x}=t^2\\ a-x=t^2(a+x)\\ x=a\cdot\frac{1-t^2}{1+t^2}\\ dx=\frac{-4at}{(t^2+1)^2}dt$

$I=-4a^2\int{\frac{t^2}{(t^2+1)^2}}dt$

Let

$t=\tan y\\ dt=\frac{dy}{\cos^2y}\\ I=-4a^2\int{\frac{\tan^2y}{(\tan^2y+1)^2}\frac{dy}{\cos^2y}}=\\ =-4a^2\int{\sin^2ydy}=-2a^2\int{(1-\cos2y)dy}=\\ =-2a^2(y-\frac{\sin2y}{4})+C$

$y=\arctan t, t=\sqrt{\frac{a-x}{a+x}}\\ y=\arctan\sqrt{\frac{a-x}{a+x}}$

then

$I=-2a^2(\arctan\sqrt{\frac{a-x}{a+x}}-\frac{\sin(2\arctan\sqrt{\frac{a-x}{a+x}})}{4})+C$