Question

$\underset{0}{\overset{1}{\int }}\sqrt{\frac{1-x}{1+x}}dx$ is equal to

• A. $\frac{\pi }{2}$
• B. $\frac{\pi }{2}-1$
• C. $\frac{\pi }{2}+1$
• D. $\pi +1$

Answer 1

The correct option is B $\frac{\pi }{2}-1$

${\int }_0^1\frac{\sqrt{1-x}}{\sqrt{1+x}}dx$

$={\int }_0^1\frac{\sqrt{1-x}\times \sqrt{1-x}}{\sqrt{1+x}\times \sqrt{1-x}}dx$

$={\int }_0^1\frac{(1-x)}{\sqrt{1-x^2}}dx$

$={\int }_0^1\frac{dx}{\sqrt{1-x^2}}-{\int }_0^1\frac{xdx}{\sqrt{1-x^2}}$

$={\int }_0^1\frac{dx}{\sqrt{1-x^2}}+\frac{1}{2}{\int }_0^1\frac{-2xdx}{\sqrt{1-x^2}}$

$={\left[{\sin }^{-1}x\right]}_0^1+\frac{1}{2}{\int }_0^1\frac{-2xdx}{\sqrt{1-x^2}}$

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

$={\left[{\sin }^{-1}x\right]}_0^1+\frac{1}{2}{\int }_0^1\frac{dt}{\sqrt{t}}$

$={\left[{\sin }^{-1}x\right]}_0^1+\frac{1}{2}{\int }_0^1{t}^{-\frac{1}{2}}dt$

$={\left[{\sin }^{-1}x\right]}_0^1+\frac{1}{2}{\left[\frac{t^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}\right]}_0^1$

$={\left[{\sin }^{-1}x\right]}_0^1+{\left[\sqrt{t}\right]}_0^1$

$={\left[{\sin }^{-1}x\right]}_0^1+{\left[\sqrt{1-x^2}\right]}_0^1$ where $t=1-x^2$

$={\sin }^{-1}1-{\sin }^{-1}0+\sqrt{1-1}-\sqrt{1-0}$

$=\frac{\pi }{2}-0-0-1$

$=\frac{\pi }{2}-1$