The value of \(\int_{0}^{1}\displaystyle\sqrt{\frac{1-x}{1+x}}\,dx\) is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(\displaystyle\frac{\pi}{2}-1\)
Multiplying the numerator and the denominator inside the root by \(1-x\):
\( \displaystyle \sqrt{\frac{1-x}{1+x}} = \frac{1-x}{\sqrt{1-x^{2}}} = \frac{1}{\sqrt{1-x^{2}}} - \frac{x}{\sqrt{1-x^{2}}} \)
The first part gives \(\sin^{-1}x\). For the second, \(\frac{d}{dx}\sqrt{1-x^{2}} = \frac{-x}{\sqrt{1-x^{2}}}\). So
\( \displaystyle \int_{0}^{1}\sqrt{\frac{1-x}{1+x}}\,dx = \left[\sin^{-1}x + \sqrt{1-x^{2}}\right]_{0}^{1} \)
\( \displaystyle = \left(\frac{\pi}{2} + 0\right) - \left(0 + 1\right) \)
\( \displaystyle = \frac{\pi}{2} - 1 \)