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Example 8 - Chapter 7 Class 12 Integrals - Part 2

Example 8 - Chapter 7 Class 12 Integrals - Part 3

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Example 8 Find the following integrals: (ii) ∫1▒𝑑π‘₯/√(2π‘₯ βˆ’ π‘₯^2 ) ∫1▒𝑑π‘₯/√(2π‘₯ βˆ’ π‘₯^2 ) Adding and subtracting 1 in denominator = ∫1▒𝑑π‘₯/√(βˆ’γ€– π‘₯γ€—^2 + 2π‘₯ + 1βˆ’1) = ∫1▒𝑑π‘₯/√(1 βˆ’ π‘₯^2 + 2π‘₯ βˆ’ 1) = ∫1▒𝑑π‘₯/√(1 βˆ’ (π‘₯^2 βˆ’ 2π‘₯ + 1) ) = ∫1▒𝑑π‘₯/√(1 βˆ’γ€– (π‘₯ βˆ’ 1)γ€—^2 ) = ∫1▒𝑑π‘₯/√(1^2 βˆ’γ€– (π‘₯ βˆ’ 1)γ€—^2 ) =sin^(βˆ’1)⁑〖(π‘₯ βˆ’ 1)/1+Cγ€— =γ€–π’”π’Šπ’γ€—^(βˆ’πŸ)⁑〖(π’™βˆ’πŸ)+π‘ͺγ€— It is of the form 𝑑π‘₯/√(π‘Ž^2 βˆ’ π‘₯^2 )=sin^(βˆ’1) π‘₯/π‘Ž+𝐢 Replacing a with 1 and π‘₯ with π‘₯βˆ’1

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