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 102 Maths Textbook Solution.

Answers (1)

Answer: $-2 \sqrt{1-x}-\frac{2}{5}(1-x)^{5 / 2}+\frac{4}{3}(1-x)^{3 / 2}+c$

Hint: to solve this question we have to use differentiate method

Given:

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

Solution:

$\text { Let } 1-x=t$

$x=1-t$

$\text { differentiating on both sides, }$

$-d x=d t$

$d x=-d t$

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

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

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

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

$I=-\left(\frac{t^{1 / 2}}{1 / 2}\right)-\int t^{3 / 2} d t+2 \frac{t^{3 / 2}}{3 / 2}$

$I=-2 \sqrt{t}-\frac{t^{3 / 2+1}}{3 / 2+1}+2 x \frac{2}{3}-t^{3 / 2}+c$

$I=-2 \sqrt{t}-\frac{t^{5 / 2}}{5 / 2}+\frac{4}{3} t^{3 / 2}+c$

$I=-2 \sqrt{1-x}-\frac{2}{5}(1-x)^{5 / 2}+\frac{4}{3}(1-x)^{3 / 2}+c$

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

'Award grace marks': CBSE's July 28 supplementary exam sparks student outrage over college admissions

anam-khan

CBSE Class 12 supplementary exam 2026 on July 28; LOC submission starts

anam-khan

CBSE OSM Evaluation Row: Class 12 revised marksheet sparks fresh debate, Vedant disputes board's explanation

Latest Question