SATHEE: Definite Integration Question 24

Definite Integration Question 24

Question 24

The value of the integral $\int_{0}^{1} \sqrt{\frac{1 - x}{1 + x}} d x$ is

(2004, 1M) (a) $\frac{π}{2} + 1$ (b) $\frac{π}{2} - 1$ (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, $t^{2} = 1 - x^{2} \Rightarrow t d t = - x d x$ ]

$= (\sin^{- 1} 1 - \sin^{- 1} 0) + [t]_{1}^{0} = π / 2 - 1$

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Definite Integration Question 24

Question 24

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