∫ dx/√(9 - 25x 2 )

(a) sin -1 (5x/3)+ c                                

(b) 1/5 sin -1 (5x/3) + c

(c) 1/6 log ((3 + 5x)/(3 -5x)) + c          

(d) 1/30 log ((3 + 5x)/(3 -5x)) + c

Integration of Square root (9 - 25 x^2) - CBSE Sample Paper - Teachoo

Question 8 - CBSE Class 12 Sample Paper for 2020 Boards - Part 2

Note : - This is same as Ex 7.4, 4 of NCERT – Chapter 7 Class 12

Check the answer here https://www.teachoo.com/5001/719/Ex-7.4--4---Integrate-1---root-9---25-x2---Class-12-NCERT/category/Ex-7.4/

 

Go Ad-free

Transcript

Question 8 โˆซ ๐‘‘๐‘ฅ/โˆš(9 โˆ’ 25๐‘ฅ^2 ) (a) sin-1 (5๐‘ฅ/3)+ c (b) 1/5 sin-1 (5๐‘ฅ/3) + c (c) 1/6 log ((3 + 5๐‘ฅ)/(3 โˆ’5๐‘ฅ)) + c (d) 1/30 log ((3 + 5๐‘ฅ)/(3 โˆ’5๐‘ฅ)) + c โˆซ1โ–’1/โˆš(9 โˆ’ 25๐‘ฅ^2 ) ๐‘‘๐‘ฅ =โˆซ1โ–’1/โˆš(25(9/25 โˆ’ ๐‘ฅ^2 ) ) ๐‘‘๐‘ฅ =โˆซ1โ–’1/(โˆš25 โˆš(9/25 โˆ’ ๐‘ฅ^2 )) ๐‘‘๐‘ฅ =1/5 โˆซ1โ–’1/โˆš(9/25 โˆ’ ๐‘ฅ^2 ) ๐‘‘๐‘ฅ =1/5 โˆซ1โ–’1/โˆš((3/5)^2 โˆ’ ๐‘ฅ^2 ) ๐‘‘๐‘ฅ It is of the form โˆซ1โ–’๐‘‘๐‘ฅ/โˆš(๐‘Ž^2 โˆ’ ๐‘ฅ^2 ) =ใ€–๐‘ ๐‘–๐‘›ใ€—^(โˆ’1)โกใ€–๐‘ฅ/๐‘Žใ€— +๐‘ โˆด Replacing ๐‘Ž by 3/5 , we get =1/5 [sin^(โˆ’1)โกใ€–๐‘ฅ/(3/5)ใ€— +๐ถ1] =๐Ÿ/๐Ÿ“ ใ€–๐’”๐’Š๐’ใ€—^(โˆ’๐Ÿ)โกใ€–๐Ÿ“๐’™/๐Ÿ‘ใ€— +๐‘ช So, (b) is the correct answer

Davneet Singh's photo - Co-founder, Teachoo

Made by

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo