Subscribe to our Youtube Channel - https://www.youtube.com/channel/UCZBx269Tl5Os5NHlSbVX4Kg

Slide27.JPG

Slide28.JPG
Slide29.JPG

  1. Chapter 7 Class 12 Integrals
  2. Serial order wise

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

About the Author

Davneet Singh's photo - Teacher, Computer Engineer, Marketer
Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.