First slide

Evaluation of definite integrals

Question

The definite integral $\int_{0}^{1} \sqrt{\frac{1 - x}{1 + x}} d x$ is equal to

Easy

A

1
B

$π$
C

$\frac{π}{2} - 1$
D

$\frac{π}{2} + \frac{1}{2}$

Solution

$\begin{aligned} I = \int_{0}^{1} \sqrt{\frac{1 - x}{1 + x}} d x = \int_{0}^{1} \frac{1 - x}{\sqrt{1 - x^{2}}} d x \\ {= (\sin^{- 1} x + \sqrt{1 - x^{2}})]}_{0}^{1} = \frac{π}{2} - 1 \end{aligned}$

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