If \(\int\displaystyle\sqrt{\frac{1-x}{1+x}}\, dx = \alpha\sqrt{1-x^2}+\beta\sin^{-1}x+C\), where C is an arbitrary constant, then which of the following are TRUE?
(A) \(\alpha = 1\)
(B) \(\alpha = -1\)
(C) \(\beta=1\)
(D) \(\beta=-1\)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A) and (C) only
For \(-1 \lt x \lt 1\), multiplying the numerator and denominator inside the root by \(1 - x\) (which is positive) rationalises the integrand:
\( \displaystyle \sqrt{\frac{1-x}{1+x}} = \sqrt{\frac{(1-x)^2}{1-x^2}} \)
\( \displaystyle = \frac{1-x}{\sqrt{1-x^2}} \)
Splitting the fraction:
\( \displaystyle \int\frac{1-x}{\sqrt{1-x^2}}\,dx = \int\frac{dx}{\sqrt{1-x^2}} - \int\frac{x}{\sqrt{1-x^2}}\,dx \)
The first integral is \(\sin^{-1}x\). For the second, let \(t = 1 - x^2\), so \(dt = -2x\,dx\):
\( \displaystyle \int\frac{x}{\sqrt{1-x^2}}\,dx = -\frac{1}{2}\int t^{-1/2}\,dt \)
\( \displaystyle = -\sqrt{t} = -\sqrt{1-x^2} \)
Hence
\( \displaystyle \int\sqrt{\frac{1-x}{1+x}}\,dx = \sqrt{1-x^2} + \sin^{-1}x + C \)
So \(\alpha = 1\) and \(\beta = 1\).
Checking the statements
(A) \(\alpha = 1\) — Correct. The coefficient of \(\sqrt{1-x^2}\) is 1.
(B) \(\alpha = -1\) — Incorrect. \(\alpha\) is 1.
(C) \(\beta=1\) — Correct. The coefficient of \(\sin^{-1}x\) is 1.
(D) \(\beta=-1\) — Incorrect. \(\beta\) is 1.
Shortcut: Differentiate the answer with \(\alpha = \beta = 1\):
\( \displaystyle \frac{d}{dx}\left(\sqrt{1-x^2} + \sin^{-1}x\right) = \frac{-x}{\sqrt{1-x^2}} + \frac{1}{\sqrt{1-x^2}} \)
\( \displaystyle = \frac{1-x}{\sqrt{1-x^2}} = \sqrt{\frac{1-x}{1+x}} \)