Question

If $I=\int \sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}dx,$ then I equals

• A. $(\sqrt{x}-2)\sqrt{1-x}+{\cos }^{-1}\sqrt{x}+C$
• B. $(\sqrt{x}+2)\sqrt{1-x}+{\sin }^{-1}\sqrt{x}+C$
• C. $(2-\sqrt{x})\sqrt{1-x}+{\cos }^{-1}\sqrt{x}+C$
• D. $(2+\sqrt{x})\sqrt{1-x}-{\sin }^{-1}\sqrt{x}+C$

Answer 1

The correct option is A $(\sqrt{x}-2)\sqrt{1-x}+{\cos }^{-1}\sqrt{x}+C$

Let $I=\int \sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}dx$
Put $\sqrt{x}=t\Rightarrow x=t^2\Rightarrow dx=2tdt$
$I=\int \sqrt{\frac{1-t}{1+t}}\left(2t\right)dt=2\int \frac{t(1-t)}{\sqrt{1-t^2}}dt=2\int \frac{t-1+1-t^2}{\sqrt{1-t^2}}dt$
$=2\int \frac{t}{\sqrt{1-t^2}}dt-2\int \frac{dt}{\sqrt{1-t^2}}+2\int \sqrt{1-t^2}dt$
$=-2\sqrt{1-t^2}-2{\sin }^{-1}t+2[\frac{t}{2}\sqrt{1-t^2}+\frac{1}{2}{\sin }^{-1}t]$
$=(t-2)\sqrt{1-t^2}-{\sin }^{-1}t+c=(\sqrt{x}-2)\sqrt{1-x}-{\sin }^{-1}\sqrt{x}+c=(\sqrt{x}-2)\sqrt{1-x}+{\cos }^{-1}\sqrt{x}+c$