Question 8 - CBSE Class 12 Sample Paper for 2020 Boards
Last updated at Oct. 22, 2019 by Teachoo

∫ 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

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/

Transcript

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

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