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Q.

$\int \sqrt{\frac{1 - x}{1 + x}} d x$ is equal to :

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a

$\sin^{- 1} x - \frac{1}{2} \sqrt{1 - x^{2}} + c$

b

$\sin^{- 1} x + \frac{1}{2} \sqrt{1 - x^{2}} + c$

c

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

d

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

answer is D.

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Detailed Solution

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Let $I = \int \sqrt{\frac{1 - x}{1 + x}} d x$

$\begin{array}{l} = \int \sqrt{\frac{(1 - x) (1 - x)}{(1 + x) (1 - x)}} \cdot d x (rationalize) \\ = \int \sqrt{\frac{{(1 - x)}^{2}}{1 - x^{2}}} d x = \int \frac{1 - x}{\sqrt{1 - x^{2}}} d x \\ = \int \frac{d x}{\sqrt{1 - x^{2}}} - \int \frac{x d x}{\sqrt{1 - x^{2}}} \\ = \sin^{- 1} x + \sqrt{1 - x^{2}} + c . \end{array}$

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