Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • NCERT
  • Choose the correct answer. (Q25)

Q25   Choose the correct answer \int \frac{dx }{\sqrt { 9x - 4x ^2 }} \: \: equals

 A) \frac{1}{9} \sin ^{-1}\left ( \frac{9x-8}{8} \right )+ C \\\\B ) \frac{1}{2} \sin ^{-1}\left ( \frac{8x-9}{9} \right )+ C \\\\ C) \frac{1}{3} \sin ^{-1}\left ( \frac{9x-8}{8} \right )+ C \\\\ D ) \frac{1}{2} \sin ^{-1}\left ( \frac{9x-8}{8} \right )+ C

Answers (1)

best_answer

The following integration can be done as

\int \frac{dx }{\sqrt { 9x - 4x ^2 }} \: \: equals
\int \frac{1}{\sqrt{-4(x^2-\frac{9}{4}x)}}= \int \frac{1}{\sqrt{-4(x^2-\frac{9}{4}x+81/64-81/64)}}dx
                                         \\= \int \frac{1}{\sqrt{-4[(x-9/8)^2-(9/8)^2]}}dx\\ =\frac{1}{2}\int \frac{1}{\sqrt{-(x-9/8)^2+(9/8)^2}}dx\\ =\frac{1}{2}[\sin^{-1}(\frac{x-9/8}{9/8})]+C\\ =\frac{1}{2}\sin^{-1}(\frac{8x-9}{9})+C

The correct option is (B)

Posted by

manish

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 OSM controversy deepens as students allege fresh errors, seek grace marks
anam-khan

CBSE to issue revised Class-12 physical marksheets through regional offices
anam-khan

NEET aspirant Aditya Mishra becomes CBSE Science topper after re-evaluation

Latest Question