Question

If $C$ is arbitrary constant then $\int \sqrt{\frac{1-x}{1+x}}dx$ is equal to

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

Answer 1

Answer: (A)

${\sin }^{-1}x+\sqrt{1-x^2}+C$

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

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

$=\int \frac{1-x}{\sqrt{1-x^2}}dx$

$=\int \frac{dx}{\sqrt{1-x^2}}-\int \frac{xdx}{\sqrt{1-x^2}}$

$=\int \frac{dx}{\sqrt{1-x^2}}+\frac{1}{2}\int \frac{-2xdx}{\sqrt{1-x^2}}$

We know that $\int \frac{dx}{\sqrt{1-x^2}}={\sin }^{-1}x$

Let $t=1-x^2\Rightarrow dt=-2xdx$

$={\sin }^{-1}x+\frac{1}{2}\int \frac{dt}{\sqrt{t}}$

$={\sin }^{-1}x+\frac{1}{2}\int t^{(-\frac{1}{2})}dt$

$={\sin }^{-1}x+\frac{1}{2}\frac{t^{(-\frac{1}{2}+1)}}{(-\frac{1}{2}+1)}+c$ where

$c$ is the constant of integration.

$={\sin }^{-1}x+\frac{1}{2}\frac{t^{\frac{1}{2}}}{\frac{1}{2}}+c$

$={\sin }^{-1}x+\sqrt{t}+c$

$={\sin }^{-1}x+\sqrt{1-x^2}+c$ where $t=1-x^2$