Question

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

• A. $\sqrt{x}\sqrt{1-x}-2\sqrt{1-x}+co{s}^{-1}\sqrt{x}+c$
• B. $\sqrt{x}\sqrt{1-x}-2\sqrt{1-x}+ta{n}^{-1}\sqrt{x}+c$
• C. $\sqrt{x}\sqrt{1-x}-2\sqrt{1-x}+si{n}^{-1}\sqrt{x}+c$
• D. $\sqrt{x}\sqrt{1-x}-2\sqrt{1-x}+se{c}^{-1}\sqrt{x}+c$

Answer 1

The correct option is A $\sqrt{x}\sqrt{1-x}-2\sqrt{1-x}+co{s}^{-1}\sqrt{x}+c$

$\int \sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}dx$
Substitute $\sqrt{x}=\sin t$
$x={\sin }^{2}t$
$dx=2\sin t\cos t$
$I=2\int \sqrt{\frac{1-\sin t}{1+\sin t}}\sin t\cos tdt$
$=2\int \sqrt{\frac{1-\sin t}{1+\sin t}\times \frac{1-\sin t}{1-\sin t}}\sin t\cos tdt$
$=2\int \frac{1-\sin t}{\cos t}\sin t\cos tdt$
$=2\int (\sin t-{\sin }^{2}t)dt$
$=2\int (\sin t-\frac{(1-\cos 2t)}{2})dt$
$=2[-\cos t-\frac{1}{2}t+\frac{\sin 2t}{4}]+C$
$=-2\cos t-t+\sin t\cos t+C$
$=-2\sqrt{1-x}-{\sin }^{-1}\sqrt{x}+\sqrt{x}\sqrt{1-x}+C$
$=-2\sqrt{1-x}+{\cos }^{-1}\sqrt{x}+\sqrt{x}\sqrt{1-x}+c$ ($\because {\sin }^{-1}x+{\cos }^{-1}x=\frac{\pi }{2}$)