Question

$\underset{0}{\overset{\mathrm{\pi }}{\int }}\sqrt{\frac{1-x}{1+x}}dx=$

(a) $\frac{\mathrm{\pi }}{2}$

(b) $\frac{\mathrm{\pi }}{2}-1$

(c) $\frac{\mathrm{\pi }}{2}+1$

(d) π + 1

Answer 1

Disclaimer: None of the given option is correct.

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ I={\int }_0^{\mathrm{\pi }}\sqrt{\frac{1-x}{1+x}}dx \\ ={\int }_0^{\mathrm{\pi }}\left[\sqrt{\frac{1-x}{1+x}}\times \frac{\sqrt{1-x}}{\sqrt{1-x}}\right]dx \\ ={\int }_0^{\mathrm{\pi }}\frac{1-x}{\sqrt{1-x^2}}dx \\ ={\int }_0^{\mathrm{\pi }}\frac{1}{\sqrt{1-x^2}}dx-{\int }_0^{\mathrm{\pi }}\frac{x}{\sqrt{1-x^2}}dx \\ \mathrm{Putting}1-x^2=t \\ \Rightarrow -2xdx=dt \\ \Rightarrow xdx=\frac{-dt}{2} \\ \mathrm{When}x\to 0;t\to 1 \\ \mathrm{and}x\to \mathrm{\pi };t\to 1-{\mathrm{\pi }}^2 \\ \therefore I={\int }_0^{\mathrm{\pi }}\frac{1}{\sqrt{1-x^2}}dx-{\int }_1^{\left(1-{\mathrm{\pi }}^2\right)}\frac{-dt}{2\sqrt{t}} \\ ={\left[{\sin }^{-1}x\right]}_0^{\mathrm{\pi }}+\frac{2}{2}{\left[\sqrt{t}\right]}_1^{\left(1-{\mathrm{\pi }}^2\right)} \\ =\left[0-0\right]+\left[\sqrt{1-{\mathrm{\pi }}^2}-\sqrt{1}\right] \\ =\sqrt{1-{\mathrm{\pi }}^2}-1 \end{gathered}$$