Question

The value of $I={\int }_0^1\sqrt{\frac{1-x}{1+x}}dx$ is

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

Answer 1

The correct option is B

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

The explanation for the correct answer.

Solve for the value of the given integral.

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

Let, $$\begin{gathered} x=\cos \left(2\theta \right) \\ dx=-2\sin \left(2\theta \right)d\theta \end{gathered}$$

$$\begin{gathered} I={\int }_{\frac{\mathrm{\pi }}{4}}^0\sqrt{\frac{1-\cos \left(2\theta \right)}{1+\cos \left(2\theta \right)}}\left(-2\sin \left(2\theta \right)d\theta \right) \\ ={\int }_0^{\frac{\mathrm{\pi }}{4}}\frac{\sin \left(\theta \right)}{\cos \left(\theta \right)}\left(2\sin \left(2\theta \right)d\theta \right)\left[\because \cos \left(2\theta \right)={\cos }^2\left(\theta \right)-{\sin }^2\left(\theta \right),{\cos }^2\left(\theta \right)+{\sin }^2\left(\theta \right)=1\right] \\ =2{\int }_0^{\frac{\mathrm{\pi }}{4}}\left(2{\sin }^2\left(\theta \right)d\theta \right)\left[\because \sin \left(2\theta \right)=2\sin \left(\theta \right)\cos \left(\theta \right)\right] \\ =2{\int }_0^{\frac{\mathrm{\pi }}{4}}\left(1-\cos \left(2\theta \right)\right)d\theta \left[\because \cos \left(2\theta \right)={\cos }^2\left(\theta \right)-{\sin }^2\left(\theta \right),{\cos }^2\left(\theta \right)+{\sin }^2\left(\theta \right)=1\right] \\ =2{\left[\theta -\frac{\sin \left(2\theta \right)}{2}\right]}_0^{\frac{\mathrm{\pi }}{4}} \\ =2\left(\frac{\mathrm{\pi }}{4}-\frac{1}{2}\right) \\ =\frac{\mathrm{\pi }}{2}-1 \end{gathered}$$

Hence, option(B) is the correct answer.