SATHEE: Indefinite Integration 1 Question 24

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) $\frac{π}{2} + 1$

(b) $\frac{π}{2} - 1$

(d) 1

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Solution:

$I = \int_{0}^{1} \sqrt{\frac{1 - x}{1 + x}} d x = \int_{0}^{1} \frac{1 - x}{\sqrt{1 - x^{2}}} d x$

$\begin{aligned} = \int_{0}^{1} \frac{1}{\sqrt{1 - x^{2}}} d x - \int_{0}^{1} \frac{x}{\sqrt{1 - x^{2}}} d x \\ = {[\sin^{- 1} x]}_{0}^{1} + \int_{1}^{0} \frac{t}{t} d t \end{aligned}$

[where, $t^{2} = 1 - x^{2} \Rightarrow t d t = - x d x$ ]

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

Indefinite Integration 1 Question 24

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

Solution:

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Indefinite Integration 1 Question 24

25. The value of the integral ∫011−x1+xdx is

Solution:

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