Question

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

Answer 1

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

Let $\sqrt{x}=cost\Rightarrow x=co{s}^{2}t\Rightarrow dx=-2costsintdt$

$I=\int \sqrt{\frac{1-cost}{1+cost}}(-2sintcost)dt$

$=\int \sqrt{\frac{2si{n}^{2}t/2}{2co{s}^{2}t/2}}(-2sintcost)dt$

$=\int \frac{sint/2}{cost/2}(-2\times 2sint/2cos1/2)costdt$

$=-4\int si{n}^{2}t/2costtdt$

$=-4\int \frac{(1-cost)}{2}costdt$

$=-4\int \frac{cost}{2}dt+4\int \frac{co{s}^{2}t}{2}dt$

$=-2sint+2\int \frac{(1-cost)}{2}dt+c$

$=-2sint+t+\frac{sint}{2}+c$

$=-2\sqrt{1+-co{s}^{2}t}+t+sintcost+c$

$=-2\sqrt{1-co{s}^{2}t}+t+\sqrt{1-co{s}^{2}t}cost+c$

Replacing cos t = $\sqrt{x}$

$I=-2\sqrt{1-x}+co{s}^{-1}\sqrt{x}+\sqrt{x}\sqrt{1-x}+c$