Indefinite Integration 1 Question 24
25. The value of the integral $\int _0^{1} \sqrt{\frac{1-x}{1+x}} d x$ is
(2004, 1M)
(a) $\frac{\pi}{2}+1$
(b) $\frac{\pi}{2}-1$
(c) -1
(d) 1
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Solution:
- $I=\int _0^{1} \sqrt{\frac{1-x}{1+x}} d x=\int _0^{1} \frac{1-x}{\sqrt{1-x^{2}}} d x$
$$ \begin{aligned} & =\int _0^{1} \frac{1}{\sqrt{1-x^{2}}} d x-\int _0^{1} \frac{x}{\sqrt{1-x^{2}}} d x \\ & =\left[\sin ^{-1} x\right] _0^{1}+\int _1^{0} \frac{t}{t} d t \end{aligned} $$
[where, $t^{2}=1-x^{2} \Rightarrow t d t=-x d x$ ]
$=\left(\sin ^{-1} 1-\sin ^{-1} 0\right)+[t] _1^{0}=\pi / 2-1$
