If $\int \sqrt\frac{1-x}{{1+x}} dx = a\sqrt{1-x^2} + \beta \sin^{-1}x + C$, 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:

Using the substitution $x = \cos\theta$, so $dx = -\sin\theta\, d\theta$:

$\int \sqrt{\frac{1-\cos\theta}{1+\cos\theta}} \cdot (-\sin\theta)\, d\theta$

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Using half-angle identities:

$1 - \cos\theta = 2\sin^2\!\left(\tfrac{\theta}{2}\right)$

$1 + \cos\theta = 2\cos^2\!\left(\tfrac{\theta}{2}\right)$

$\sqrt{\frac{2\sin^2(\theta/2)}{2\cos^2(\theta/2)}} = \frac{\sin(\theta/2)}{\cos(\theta/2)} = \tan\!\left(\tfrac{\theta}{2}\right)$

Also, $\sin\theta = 2\sin(\theta/2)\cos(\theta/2)$

$= \int \frac{\sin(\theta/2)}{\cos(\theta/2)} \cdot \big(-2\sin(\theta/2)\cos(\theta/2)\big)\, d\theta$

$= \int -2\sin^2\!\left(\tfrac{\theta}{2}\right)\, d\theta$

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Using $\sin^2(\theta/2) = \dfrac{1 - \cos\theta}{2}$:

$= \int -(1-\cos\theta)\, d\theta$

$= \int(-1 + \cos\theta)\, d\theta$

$= -\theta + \sin\theta + C$

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Since $x = \cos\theta$:

$\theta = \cos^{-1}x$

$\sin\theta = \sqrt{1 - x^2}$

$= -\cos^{-1}x + \sqrt{1-x^2} + C$

Using $\cos^{-1}x = \dfrac{\pi}{2} - \sin^{-1}x$:

$= -\left(\frac{\pi}{2} - \sin^{-1}x\right) + \sqrt{1-x^2} + C$

$= \sqrt{1-x^2} + \sin^{-1}x + \left(C - \frac{\pi}{2}\right)$

$= \sqrt{1-x^2} + \sin^{-1}x + C$

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Comparing with $\alpha\sqrt{1-x^2} + \beta\sin^{-1}x + C$:

$\alpha = 1$

$\beta = 1$

Therefore (A) and (C) are TRUE.