Question 8 โซ ๐๐ฅ/โ(9 โ 25๐ฅ^2 ) (a) sin-1 (5๐ฅ/3)+ c (b) 1/5 sin-1 (5๐ฅ/3) + c (c) 1/6 log ((3 + 5๐ฅ)/(3 โ5๐ฅ)) + c (d) 1/30 log ((3 + 5๐ฅ)/(3 โ5๐ฅ)) + c
โซ1โ1/โ(9 โ 25๐ฅ^2 ) ๐๐ฅ
=โซ1โ1/โ(25(9/25 โ ๐ฅ^2 ) ) ๐๐ฅ
=โซ1โ1/(โ25 โ(9/25 โ ๐ฅ^2 )) ๐๐ฅ
=1/5 โซ1โ1/โ(9/25 โ ๐ฅ^2 ) ๐๐ฅ
=1/5 โซ1โ1/โ((3/5)^2 โ ๐ฅ^2 ) ๐๐ฅ
It is of the form
โซ1โ๐๐ฅ/โ(๐^2 โ ๐ฅ^2 ) =ใ๐ ๐๐ใ^(โ1)โกใ๐ฅ/๐ใ +๐
โด Replacing ๐ by 3/5 , we get
=1/5 [sin^(โ1)โกใ๐ฅ/(3/5)ใ +๐ถ1]
=๐/๐ ใ๐๐๐ใ^(โ๐)โกใ๐๐/๐ใ +๐ช
So, (b) is the correct answer
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
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