Go To Your Study Dashboard

Get Answers to all your Questions

header-bg

Explain Solution R.D.Sharma Class 12 Chapter 18 Indefinite Integrals Exercise Revision Exercise Question 105 Maths Textbook Solution.

Answers (1)

Answer: $\sqrt{1-x^{2}}\left(\frac{x}{2}-1\right)-\frac{1}{2} \sin ^{-1}(x)+C$

Hint: to solve this equation, we have to assume u as $\cos \Theta$

Given: $\int x \sqrt{\frac{1-x}{1+x}} d x$

Solution:

$I=\int x \sqrt{\frac{1-x}{1+x}} d x$

$I=\int x \sqrt{\frac{(1-x)(1-x)}{(1+x)(1-x)}} d x$

$I=\int x \frac{(1-x)}{\sqrt{1-x^{2}}} d x$

$I=\int \frac{x-x^{2}}{\sqrt{1-x^{2}}} d x$

$I=\int \frac{x-x^{2}-1+1}{\sqrt{1-x^{2}}} d x$

$I=\int \frac{-x^{2}+1}{\sqrt{1-x^{2}}} d x+\int \frac{x-1}{\sqrt{1-x^{2}}} d x$

$I=\int \sqrt{1-x^{2}} d x+\int \frac{x}{\sqrt{1-x^{2}}} d x-\int \frac{1}{\sqrt{1-x^{2}}} d x$

$I=\frac{x}{2} \sqrt{1-x^{2}}+\frac{1}{2} \sin ^{-1}(x)+C_{1}-\sqrt{1-x^{2}}+C_{2}-\sin ^{-1}(x)$

$+C_{3}\left[\because \int \frac{x}{\sqrt{1-x^{2}}} d x=-\sqrt{1-x^{2}}+C_{2}\right]$

$I=\sqrt{1-x^{2}}\left(\frac{x}{2}-1\right)-\frac{1}{2} \sin ^{-1}(x)+C$

Posted by

infoexpert21

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

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

anam-khan

CBSE Board Exams 2025: Online portal for availing exemptions by CWSN students opens

anam-khan

Board Exams: States agree to equivalence; no question paper ‘jumbling’ from next year, says PARAKH CEO

Latest Question