Question

$\int \frac{\sqrt{1-\sqrt{x}}}{\sqrt{1+\sqrt{x}}}dx$ is equal to
(where $C,C_1$ are integration constant)

• A. $\sqrt{1-x}(\sqrt{x}-2)-{\sin }^{-1}\sqrt{x}+C$
• B. $\sqrt{1-x^2}(x-2)-{\sin }^{-1}\sqrt{x}+C$
• C. $\sqrt{1-x}(x+1)+{\cos }^{-1}\sqrt{x}+C_1$
• D. $\sqrt{1-x}(\sqrt{x}-2)+{\cos }^{-1}\sqrt{x}+C_1$

Answer 1

Answer: (A), (D)

$\sqrt{1-x}(\sqrt{x}-2)+{\cos }^{-1}\sqrt{x}+C_1$

$I=\int \frac{\sqrt{1-\sqrt{x}}}{\sqrt{1+\sqrt{x}}}dx$
Putting $x=t^2$ and $dx=2tdt$, we get
$I=2\int \sqrt{\frac{1-t}{1+t}}\cdot tdt\Rightarrow I=2\int \frac{t(1-t)}{\sqrt{1-t^2}}dt\Rightarrow I=\int (\frac{2t}{\sqrt{1-t^2}}+\frac{-2t^2}{\sqrt{1-t^2}})dt\Rightarrow I=\int \frac{2t}{\sqrt{1-t^2}}dt+2\int \frac{(1-t^2)-1}{\sqrt{1-t^2}}dt\Rightarrow I=-\int \frac{-2t}{\sqrt{1-t^2}}dt+2\int \sqrt{1-t^2}dt-2\int \frac{1}{\sqrt{1-t^2}}dt\Rightarrow I=-2\sqrt{1-t^2}+2\times \frac{1}{2}[t\sqrt{1-t^2}+{\sin }^{-1}t]-2{\sin }^{-1}t+C\Rightarrow I=-2\sqrt{1-t^2}+t\sqrt{1-t^2}-{\sin }^{-1}t+C\therefore I=\sqrt{1-x}(\sqrt{x}-2)-{\sin }^{-1}\sqrt{x}+C\Rightarrow I=\sqrt{1-x}(\sqrt{x}-2)-(\frac{\pi }{2}-{\cos }^{-1}\sqrt{x})+C\therefore I=\sqrt{1-x}(\sqrt{x}-2)+{\cos }^{-1}\sqrt{x}+C_1[\because {\sin }^{-1}x+{\cos }^{-1}=\frac{\pi }{2}]$