∫ 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

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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/

 

  1. Class 12
  2. Solutions of Sample Papers and Past Year Papers - for Class 12 Boards

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
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.