∫ 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/

 

Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class


Transcript

Question 8 ∫ 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 ∫1▒1/√(9 − 25𝑥^2 ) 𝑑𝑥 =∫1▒1/√(25(9/25 − 𝑥^2 ) ) 𝑑𝑥 =∫1▒1/(√25 √(9/25 − 𝑥^2 )) 𝑑𝑥 =1/5 ∫1▒1/√(9/25 − 𝑥^2 ) 𝑑𝑥 =1/5 ∫1▒1/√((3/5)^2 − 𝑥^2 ) 𝑑𝑥 It is of the form ∫1▒𝑑𝑥/√(𝑎^2 − 𝑥^2 ) =〖𝑠𝑖𝑛〗^(−1)⁡〖𝑥/𝑎〗 +𝑐 ∴ Replacing 𝑎 by 3/5 , we get =1/5 [sin^(−1)⁡〖𝑥/(3/5)〗 +𝐶1] =𝟏/𝟓 〖𝒔𝒊𝒏〗^(−𝟏)⁡〖𝟓𝒙/𝟑〗 +𝑪 So, (b) is the correct answer

Ask a doubt
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 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.