The value of ∫01 1−x1+xdx, is - Infinity Learn by Sri Chaitanya

A
$\frac{π}{2} + 1$
B
$\frac{π}{2} - 1$
C
$- 1$
D
1

Solution:

We have,

$\begin{array}{l} = \int_{0}^{1} \sqrt{\frac{1 - x}{1 + x}} d x \\ = \int_{0}^{1} \frac{1 - x}{\sqrt{1 - x^{2}}} d x = \int_{0}^{1} \frac{1}{\sqrt{1 - x^{2}}} d x - \int_{0}^{1} \frac{x}{\sqrt{1 - x^{2}}} d x \\ = {[\sin^{- 1} x]}_{0}^{1} + {[\sqrt{1 - x^{2}}]}_{0}^{1} = π / 2 - 1 \end{array}$

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

Solution:

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