Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • School
  • Please solve RD Sharma Class 12 Chapter 19 Definite Integrals Exercise 19.4 (a) Question 2 maths textbook solution.

Please solve RD Sharma Class 12 Chapter 19 Definite Integrals Exercise 19.4 (a) Question 2 maths textbook solution.

Answers (1)

Answer :  0

Given : \int_{0}^{2 \pi} \log (\sec x+\tan x) d x

Hint : You must know about the trignometric identities

Solution : I=\int_{0}^{2 \pi} \log (\sec x+\tan x) d x-------(1)

We know that

\begin{aligned} &\int_{0}^{2 \pi} f(x) d x=\int_{0}^{2 \pi} f(2 \pi-x) d x \\ &I=\int_{0}^{2 \pi} \log (\sec (2 \pi-x)+\tan (2 \pi-x)) d x \\ &I=\int_{0}^{2 \pi} \log (\sec x-\tan x) d x \quad-------(2)[\sec (2 \pi-x)=\sec x, \tan (2 \pi-x)=-\tan x] \end{aligned}

Add (1) and (2)

\begin{aligned} &2 I=\int_{0}^{2 \pi} \log (\sec x+\tan x)+\log (\sec x-\tan x) d x \\ &2 I=\int_{0}^{2 \pi} \log (1) d x\left[\tan ^{2} x+1=\sec ^{2} x\right] \\ &2 I=0[\log (1)=0] \\ &I=0 \end{aligned}

Posted by

infoexpert23

View full answer

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support
cuet_ads

Similar Questions

Trending Articles/News

anam-khan

CBSE extends single girl child scholarship registration deadline till November 20
anam-khan

CBSE reminds schools to submit registration data of Class 9, 11 students by October 16

Latest Question