∫ 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/

 

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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

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.