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  4. The value of the integral displaystyle ∫ 01 √( 1 - x/1 + x) d x is

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Q. The value of the integral $\displaystyle \int _{0}^{1} \sqrt{\frac{ 1 - x}{1 + x}} d x$ is

NTA AbhyasNTA Abhyas 2020Integrals

Solution:

$I=\displaystyle \int _{0}^{1}\left[\right. \sqrt{\frac{1 - x}{1 + x}} \times \frac{\sqrt{1 - x}}{\sqrt{1 - x}} \left]\right. \, \, dx \, \left(\right. rationalising \, the \, denominator \left.\right)$
$=\displaystyle \int _{0}^{1} \frac{ 1 - x}{\sqrt{ 1 - x^{2}}} d x \\ =\displaystyle \int _{0}^{1} \frac{1}{\sqrt{ 1 - x^{2}}} d x -\displaystyle \int _{0}^{1} \frac{x \, d x}{\sqrt{ 1 - x^{2}}}$
$\Rightarrow \, \, I = \left[\right. sin^{ - 1} x \left]\right._{0}^{1} - \frac{1}{2} \displaystyle \int _{0}^{1} \frac{ 2 x}{\sqrt{ 1 - x^{2}}} d x$
$= \left[\right. \frac{\pi }{2} - 0 \left]\right. + \frac{1}{2} \left[\right. 2 \sqrt{ 1 - x^{2}} \left]\right._{0}^{1}$
$=\frac{\pi }{2}+\left[\right. 0 - 1 \left]\right. \\ =\frac{\pi }{2} - 1$