Question
Evaluate the following intregals:
$\int\sqrt{\frac{1-\text{x}}{1+\text{x}}}\text{dx}$

Answer

 Let  $\text{I}=\int\sqrt{\frac{1-\text{x}}{1+\text{x}}}\text{dx}$
$=\int\sqrt{\frac{(1-\text{x})(1-\text{x})}{(1+\text{x})(1-\text{x})}}\text{dx}$
$=\int\Big(\frac{1-\text{x}}{\sqrt{1-\text{x}^2}}\Big)\text{dx}$
$=\int\frac{\text{dx}}{\sqrt{1-\text{x}^2}}-\int\frac{\text{x dx}}{\sqrt{1-\text{x}^2}}$
Putting $1-\text{x}^2=\text{t}$
$\Rightarrow-2\text{x}\text{ dx}=\text{dt}$
$\Rightarrow\text{x dx}=-\frac{\text{dt}}{2}$
Then
$\text{I}=\int\frac{\text{dx}}{\sqrt{1-\text{x}^2}}+\frac{1}{2}\int\frac{\text{dt}}{\sqrt{\text{t}}}$
$=\sin^{-1}(\text{x})+\frac{1}{2}\times2\sqrt{\text{t}}+\text{C}$
$=\sin^{-1}(\text{x})+\sqrt{1-\text{x}^2}+\text{C}$

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