∫ 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/

 


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

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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, Social Science, Physics, Chemistry, Computer Science at Teachoo.