First slide

Evaluation of definite integrals

Question

$\int_{0}^{π / 2} \frac{5 \tan x - 3 \cot x}{\tan x + \cot x} d x =$

Moderate

A

$\frac{π}{2}$
B

$\frac{π}{3}$
C

$\frac{π}{4}$
D

$\frac{π}{8}$

Solution

$\begin{array}{l} I = \int_{0}^{π / 2} \frac{5 \tan x - 3 \cot x}{\tan x + \cot x} d x - - - - - - (1) \\ I = \int_{0}^{π / 2} \frac{5 \tan (\frac{π}{2} - x) - 3 \cot (\frac{π}{2} - x)}{\tan (\frac{π}{2} - x) + \cot (\frac{π}{2} - x)} d x \end{array}$
$\begin{array}{l} I = \int_{0}^{π / 2} \frac{5 \cot x - 3 \tan x}{\cot x + \tan x} d x - - - - (2) \\ (1) + (2) \Rightarrow 2 I = \int_{0}^{π / 2} \frac{2 \tan x + 2 \cot x}{\cot x + \tan x} d x = \int_{0}^{π / 2} 2 d x \\ = 2 \int_{0}^{π / 2} (1) d x = 2 (x)_{0}^{π / 2} = 2 \frac{π}{2} = π \\ \Rightarrow I = \frac{π}{2} \end{array}$

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