Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • School
  • Provide solution for RD Sharma maths class 12 chapter 19 Definite Integrals exercise Multiple choice question 10

Provide solution for RD Sharma maths class 12 chapter 19 Definite Integrals exercise Multiple choice question 10

Answers (1)

Answer:

\frac{\pi}{2}-1

Given:

\int_{0}^{1} \sqrt{\frac{1-x}{1+x}} d x

Hint:

Explanation:  

Let

\begin{aligned} &I=\int_{0}^{1} \sqrt{\frac{1-x}{1+x}} d x \\\\ &=\int_{0}^{1} \sqrt{\frac{1-x}{1+x} \times \frac{1-x}{1-x}} d x \end{aligned}

\begin{aligned} &=\int_{0}^{1} \sqrt{\frac{(1-x)^{2}}{1-x^{2}}} d x \\\\ &=\int_{0}^{1} \frac{1-x}{\sqrt{1-x^{2}}} d x \end{aligned}

\begin{aligned} &=\int_{0}^{1} \frac{1}{\sqrt{1-x^{2}}} d x-\int_{0}^{1} x\left(1-x^{2}\right)^{-1 / 2} d x \\\\ &=\int_{0}^{1} \frac{1}{\sqrt{1-x^{2}}} d x+\int_{0}^{1}\left(1-x^{2}\right)^{-1 / 2}(-2 x) d x \end{aligned}

\begin{aligned} &=\left[\sin ^{-1} x\right]_{0}^{1}+\frac{1}{2}\left[\frac{\left(1-x^{2}\right)^{1 / 2}}{1 / 2}\right]_{0}^{1} \\\\ &=\left[\sin ^{-1}(1)-\sin ^{-1}(0)\right]+\frac{1}{2}\left[0-\frac{1}{1 / 2}\right] \end{aligned}

\begin{aligned} &=\left(\frac{\pi}{2}-0\right)+\frac{1}{2}(-2) \\\\ &=\frac{\pi}{2}-1 \end{aligned}

 

Posted by

infoexpert26

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 SARAS 7.0: Affiliation application deadline extended to September 30 for 2027-28 session
anam-khan

CBSE Class 12 supplementary exam 2026 on July 28; major subjects, timings, exam day guidelines
anam-khan

CBSE Class 12 supplementary practical exams from July 29; check exam guidelines, rules

Latest Question