If $\int\sqrt{\frac{1-x}{1+x}} dx = α\sqrt{1-x^2}+β\sin^{-1}x+C$, Where C is an arbitrary constant, then which of the following are TRUE?
(A) $α = 1$
(B) $α = -1$
(C) $β=1$
(D) $β=-1$
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
Evaluate: $\int \sqrt{\frac{1-x}{1+x}}\,dx$
$\sqrt{\frac{1-x}{1+x}} = \sqrt{\frac{(1-x)^2}{1-x^2}} = \frac{1-x}{\sqrt{1-x^2}} \ \text{for } -1 < x < 1$
$\Rightarrow \int \sqrt{\frac{1-x}{1+x}}\,dx = \int \frac{1}{\sqrt{1-x^2}}\,dx - \int \frac{x}{\sqrt{1-x^2}}\,dx$
$= \sin^{-1}x - (-\sqrt{1-x^2}) + C = \sqrt{1-x^2} + \sin^{-1}x + C$
Compare with $\alpha\sqrt{1-x^2} + \beta\sin^{-1}x + C$
$\Rightarrow \alpha = 1,\ \beta = 1$
Final Answer:
(A) $\alpha = 1$ and (C) $\beta = 1$