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Ex 7.4, 25 - Integration dx / root 9x - 4x^2 equals (A) 1/9 sin-1

Ex 7.4, 25 - Chapter 7 Class 12 Integrals - Part 2
Ex 7.4, 25 - Chapter 7 Class 12 Integrals - Part 3

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Transcript

Ex 7.4, 25 ∫1▒𝑑π‘₯/√(9π‘₯ βˆ’ 4π‘₯^2 ) equals A. 1/9 sinβˆ’1 ((9π‘₯ βˆ’ 8)/8) + C B. 1/9 sinβˆ’1 ((8π‘₯ βˆ’ 9)/9) + C C. 1/3 sinβˆ’1 ((9π‘₯ βˆ’ 8)/8) + C D. 1/2 sinβˆ’1 ((9π‘₯ βˆ’ 8)/8) + C ∫1▒𝑑π‘₯/√(9π‘₯ βˆ’ 4π‘₯^2 ) =∫1▒𝑑π‘₯/√(βˆ’4(π‘₯^2 βˆ’ 9/4 π‘₯) ) =∫1▒𝑑π‘₯/√(βˆ’4(π‘₯^2 βˆ’ 2(π‘₯) (9/8)) ) (Taking βˆ’4 common) =∫1▒𝑑π‘₯/√(βˆ’4[π‘₯^2 βˆ’ 2(π‘₯) (9/8) + (9/8)^2βˆ’ (9/8)^2 ] ) =∫1▒𝑑π‘₯/√(βˆ’4[(π‘₯ βˆ’ 9/8)^2βˆ’ (9/8)^2 ] ) =∫1▒𝑑π‘₯/√(4[(9/8)^2 βˆ’ (π‘₯ βˆ’ 9/8)^2 ] ) =∫1▒𝑑π‘₯/(√4 √((9/8)^2 βˆ’ (π‘₯ βˆ’ 9/8)^2 )) =1/2 ∫1▒𝑑π‘₯/√((9/8)^2 βˆ’ (π‘₯ βˆ’ 9/8)^2 ) It is of form ∫1▒𝑑π‘₯/√(π‘Ž^2 βˆ’ π‘₯^2 ) =sin^(βˆ’1)⁑〖π‘₯/π‘Žγ€— +𝐢1 ∴ Replacing π‘₯ by (π‘₯βˆ’ 9/8) and π‘Ž by 9/8 , we get =1/2 [sin^(βˆ’1)⁑〖(π‘₯ βˆ’ 9/8)/(9/8)γ€— +𝐢1] =1/2 sin^(βˆ’1)⁑[(π‘₯ βˆ’ 9/8)/(9/8)] +𝐢 =1/2 sin^(βˆ’1)⁑〖((8π‘₯ βˆ’ 9)/8)/(9/8)γ€— +𝐢 =1/2 sin^(βˆ’1)⁑〖(8π‘₯ βˆ’ 9)/9γ€— +𝐢 ∴ Option B is 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 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.