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Class 12
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∫ dx/√(9 - 25x 2 )

(a) sin -1 (5x/3)+ cΒ  Β  Β  Β  Β  Β  Β  Β  Β  Β  Β  Β  Β  Β  Β  Β Β 

(b) 1/5 sin -1 (5x/3) + c

(c) 1/6 log ((3 + 5x)/(3 -5x)) + cΒ  Β  Β  Β  Β Β 

(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/

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Transcript

Question 8 ∫ 𝑑π‘₯/√(9 βˆ’ 25π‘₯^2 ) (a) sin-1 (5π‘₯/3)+ c (b) 1/5 sin-1 (5π‘₯/3) + c (c) 1/6 log ((3 + 5π‘₯)/(3 βˆ’5π‘₯)) + c (d) 1/30 log ((3 + 5π‘₯)/(3 βˆ’5π‘₯)) + c ∫1β–’1/√(9 βˆ’ 25π‘₯^2 ) 𝑑π‘₯ =∫1β–’1/√(25(9/25 βˆ’ π‘₯^2 ) ) 𝑑π‘₯ =∫1β–’1/(√25 √(9/25 βˆ’ π‘₯^2 )) 𝑑π‘₯ =1/5 ∫1β–’1/√(9/25 βˆ’ π‘₯^2 ) 𝑑π‘₯ =1/5 ∫1β–’1/√((3/5)^2 βˆ’ π‘₯^2 ) 𝑑π‘₯ It is of the form ∫1▒𝑑π‘₯/√(π‘Ž^2 βˆ’ π‘₯^2 ) =〖𝑠𝑖𝑛〗^(βˆ’1)⁑〖π‘₯/π‘Žγ€— +𝑐 ∴ Replacing π‘Ž by 3/5 , we get =1/5 [sin^(βˆ’1)⁑〖π‘₯/(3/5)γ€— +𝐢1] =𝟏/πŸ“ γ€–π’”π’Šπ’γ€—^(βˆ’πŸ)β‘γ€–πŸ“π’™/πŸ‘γ€— +π‘ͺ So, (b) is the correct answer

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Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.