Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • School
  • Need solution for RD Sharma Maths Class 12 Chapter 18 Indefinite Integrals Excercise 18 .14 Question 9

Need solution for RD Sharma Maths Class 12 Chapter 18 Indefinite Integrals Excercise 18.14 Question 9

Answers (1)

Answer:  -\log \left|(2-x)+\sqrt{(2-x)^{2}-1}\right|+c

Hint: To solve this integral, use special integral formula.

Given:   \int \frac{1}{\sqrt{(2-x)^{2}-1}} d x \\

Solution:

Let

 \begin{aligned} &I=\int \frac{1}{\sqrt{(2-x)^{2}-1}} d x \\ \end{aligned}

\text { Put } 2-x=t \Rightarrow-d x=d t \Rightarrow d x=-d t \text { then }

\begin{aligned} &I=\int \frac{1}{\sqrt{t^{2}-1}}(-d t)=-\int \frac{1}{\sqrt{t^{2}-1^{2}}} d t \\ & \end{aligned}

=-\log \left|t+\sqrt{t^{2}-1}\right|+c \quad\quad\quad\quad\quad\quad\left[\because \int \frac{1}{\sqrt{x^{2}-a^{2}}} d x=\log \left|x+\sqrt{x^{2}-a^{2}}\right|+c\right]

=-\log \left|(2-x)+\sqrt{(2-x)^{2}-1}\right|+c \quad \quad \quad \quad \quad[\because t=2-x]

Posted by

infoexpert27

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 Single Girl Child Scholarship 2024: Registration extended till January 10; eligibility criteria, amount
anam-khan

CBSE wants international boards reined in; letter to education ministry seeks directions for AIU
anam-khan

PM Modi's Pariksha Pe Charcha 2025 to be held next month, registrations open till January 14

Latest Question