First slide

Evaluation of definite integrals

Question

The equal to integral $\int_{0}^{1} \log (\sqrt{1 - x} + \sqrt{1 + x}) d x$ is

Moderate

A

$\frac{1}{2} (\log 2 - \frac{1}{2} + \frac{π}{4})$
B

$\frac{1}{2} (\log 2 - 1 + \frac{π}{2})$
C

$\frac{1}{3} (\log 4 - 1 + \frac{π}{8})$
D

$\frac{1}{4} (\log 3 - 1 + \frac{π}{2})$

Solution

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

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App