First slide

Evaluation of definite integrals

Question

The value $I = \int_{0}^{π / 2} \frac{1}{1 + \cot x} d x$ is

Easy

A

$π$
B

$π / 2$
C

$π / 4$
D

$π / 3$

Solution

Let $f (x) = \frac{1}{1 + \cot x} = \frac{\sin x}{\sin x + \cos x},$
$\begin{array}{l} f (π / 2 - x) = \frac{\sin (π / 2 - x)}{\sin (π / 2 - x) + \cos (π / 2 - x)} \\ = \frac{\cos x}{\cos x + \sin x} \\ Considering 2 I = \int_{0}^{π / 2} f (x) d x + \int_{0}^{π / 2} f (π / 2 - x) d x \\ = \int_{0}^{π / 2} \frac{\sin x}{\cos x + \sin x} d x + \int_{0}^{π / 2} \frac{\cos x}{\cos x + \sin x} d x \end{array}$
$= \int_{0}^{π / 2} \frac{\sin x + \cos x}{\cos x + \sin x} d x = \int_{0}^{π / 2} 1 \cdot d x = \frac{π}{2}$
So $I = π / 4$

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