Question

Solve:

$\int \sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}dx$

• A. $\sqrt{1-x}-\frac{\sqrt{x(1-x)}+{\cos }^{-1}\sqrt{x}}{4}+c$
  
• B. $\sqrt{1-x}-\frac{\sqrt{x(1-x)}-{\cos }^{-1}\sqrt{x}}{2}+c$
  
• C. $-\sqrt{1-x}+\frac{\sqrt{x(1-x)}+{\cos }^{-1}\sqrt{x}}{2}+c$
  
• D. $-\sqrt{1-x}+\frac{\sqrt{1-x}-{\cos }^{-1}\sqrt{x}}{2}+c$
  

Answer 1

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

$I=\int \sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}dx$

Let $x={\cos }^{2}2m$ $\Rightarrow dx=-2\cos 2m\sin 2mdm$

$\Rightarrow I=\int \sqrt{\frac{1-\cos 2m}{1+\cos 2m}}\times (-\sin 4m)dm$

$=-\int \sqrt{\frac{{\sin }^{2}m}{{\cos }^{2}m}}\sin mdm$

$=-\int \frac{\sin m}{\cos m}\sin 4mdm$

$=-4\int \sin m\sin m\times \cos 2mdm$

$=-4\int {\sin }^{2}m\cos 2mdm$

$=-\frac{4}{2}\int (1-\cos 2m)\cos 2mdm$

$=-2\int \cos 2mdm+2\int {\cos }^{2}2mdm$

$=-\sin 2m+\int (\cos 4m+1)dm$

$=-\sin 2m+\frac{1}{4}\left(\sin 4m\right)+m+c$

$=-\sin 2m+\frac{\sin 4m}{4}+m+c$

$=-\sqrt{1-x}+\frac{\sqrt{x(1-x)}+{\cos }^{-1}\sqrt{x}}{2}+c$