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  1. Chapter 7 Class 12 Integrals (Term 2)
  2. Serial order wise

Transcript

Example 10 Find the following integrals: (ii) ∫1β–’(π‘₯ + 3)/√(5 βˆ’ 4π‘₯ βˆ’ π‘₯^2 ) 𝑑π‘₯ We can write numerator as π‘₯+3= A𝑑/𝑑π‘₯ [βˆ’π‘₯^2βˆ’4π‘₯+5]+ B π‘₯+3= A[βˆ’2π‘₯βˆ’4]+ B π‘₯+3=βˆ’2𝐴π‘₯βˆ’4A+B Finding A & B Comparing coefficient of π‘₯ π‘₯=βˆ’2𝐴π‘₯ 1 =βˆ’2A A=(βˆ’1)/( 2) Comparing constant term 3=βˆ’4A+B 3=βˆ’4((βˆ’1)/( 2))+B 3=2+B B=3βˆ’2=1 Now, we know that π‘₯+3=A[βˆ’2π‘₯βˆ’4]+B π‘₯+3=(βˆ’1)/( 2) [βˆ’2π‘₯βˆ’4]+1 Now, our equation is ∫1β–’(π‘₯+3)/√(5 βˆ’ 4π‘₯ + π‘₯^2 ).𝑑π‘₯=∫1β–’(1/2 [βˆ’2π‘₯ βˆ’ 4]+1)/√(5 βˆ’ 4π‘₯ βˆ’ π‘₯^2 ).𝑑π‘₯ =∫1β–’((βˆ’1)/2 [βˆ’2π‘₯ βˆ’ 4]+1)/√(5 βˆ’ 4π‘₯ + π‘₯^2 ) 𝑑π‘₯+∫1β–’1/√(5 βˆ’ 4π‘₯ βˆ’ π‘₯^2 ) 𝑑π‘₯ =(βˆ’1)/2 ∫1β–’(βˆ’2π‘₯ βˆ’ 4)/√(5 βˆ’ 4π‘₯ + π‘₯^2 ) 𝑑π‘₯+∫1β–’1/√(5 βˆ’ 4π‘₯ βˆ’ π‘₯^2 ) 𝑑π‘₯ Solving I1 I1 = (βˆ’1)/2 ∫1β–’γ€–(βˆ’2π‘₯ βˆ’ 4)/√(βˆ’π‘₯^2 βˆ’ 4π‘₯ + 5).𝑑π‘₯γ€— Let t = βˆ’π‘₯^2 βˆ’ 4π‘₯ + 5 Differentiating both sides w.r.t. π‘₯ βˆ’2π‘₯ βˆ’ 4 =𝑑𝑑/𝑑π‘₯ 𝑑π‘₯=𝑑𝑑/(βˆ’ 2π‘₯ βˆ’ 4) Now, I1 = (βˆ’1)/( 2) ∫1β–’γ€–(βˆ’2π‘₯ βˆ’ 4)/√(βˆ’π‘₯^2 βˆ’ 4π‘₯ + 5).𝑑π‘₯γ€— Putting the values of (βˆ’π‘₯^2βˆ’4π‘₯+5) and 𝑑π‘₯ I1 = (βˆ’1)/( 2) ∫1β–’γ€–(βˆ’2π‘₯ βˆ’ 4)/βˆšπ‘‘.𝑑𝑑/((βˆ’2π‘₯ βˆ’4) )γ€— I1 = (βˆ’1)/( 2) ∫1β–’γ€–1/βˆšπ‘‘.𝑑𝑑〗 I1 = (βˆ’1)/( 2) ∫1β–’γ€–1/(𝑑)^(1/2) .𝑑𝑑〗 I1 = (βˆ’1)/( 2) ∫1β–’γ€–(𝑑)^((βˆ’1)/2) 𝑑𝑑〗 I1 = (βˆ’1)/( 2) 𝑑^((βˆ’1)/( 2) + 1)/(((βˆ’1)/( 2) + 1) )+𝐢1 I1 = (βˆ’1)/( 2) 𝑑^(1/( 2) )/((1/2) )+𝐢1 I1 = βˆ’ 𝑑^(1/( 2) )+𝐢1 I1 = βˆ’βˆšπ‘‘+𝐢1 Putting back 𝑑=π‘₯^2βˆ’4π‘₯+5) I1 = βˆ’βˆš(βˆ’π‘₯^2βˆ’4π‘₯+5)+𝐢1 Solving π‘°πŸ I2 = ∫1β–’γ€–1/√(5 βˆ’ 4π‘₯ βˆ’ π‘₯^2 ).𝑑π‘₯γ€— I2 = ∫1β–’γ€–1/√(βˆ’(π‘₯^2 + 4π‘₯ βˆ’ 5) ) 𝑑π‘₯γ€— I2 = ∫1β–’γ€–1/√(βˆ’[π‘₯^2 + 2(2)(π‘₯) βˆ’ 5] ) 𝑑π‘₯γ€— I2 = ∫1β–’γ€–1/√(βˆ’[π‘₯^2 + 2(2)(π‘₯) + (2)^2 βˆ’ (2)^2 βˆ’ 5] ) 𝑑π‘₯γ€— I2 = ∫1β–’γ€–1/√(βˆ’[(π‘₯ + 2)^2 βˆ’ 4 βˆ’ 5] ) 𝑑π‘₯γ€— I2 = ∫1β–’γ€–1/√(βˆ’[(π‘₯ + 2)^2 βˆ’ 9] ) 𝑑π‘₯γ€— I2 = ∫1β–’γ€–1/√(9 βˆ’ (π‘₯ + 2)^2 ) 𝑑π‘₯γ€— I2 = ∫1β–’γ€–1/√((3)^2 βˆ’ (π‘₯ + 2)^2 ) 𝑑π‘₯γ€— It is of form ∫1β–’γ€–1/√(π‘Ž^2 βˆ’ π‘₯^2 ).𝑑π‘₯=〖𝑠𝑖𝑛〗^(βˆ’1)⁑〖π‘₯/π‘Ž+𝐢2γ€— γ€— Replacing π‘₯ by (π‘₯+2) and a by 3, we get I2 = sin^(βˆ’1)⁑〖((π‘₯ + 2)/3)+𝐢2γ€— Now, Putting values of I1 and I2 in (1) ∫1β–’γ€–(π‘₯ + 3)/√(5 βˆ’ 4π‘₯ βˆ’ π‘₯^2 ).𝑑π‘₯=(βˆ’1)/( 2) ∫1β–’(βˆ’ 2π‘₯ βˆ’ 4)/√(5 βˆ’ 4π‘₯ βˆ’ π‘₯^2 )γ€—+∫1β–’1/√(5 βˆ’ 4π‘₯ βˆ’ π‘₯^2 ) 𝑑π‘₯ =βˆ’βˆš(5 βˆ’ 4π‘₯ βˆ’ π‘₯^2 )+𝐢1+sin^(βˆ’1)⁑((π‘₯ + 2)/3)+𝐢2 =βˆ’βˆš(πŸ“ βˆ’ πŸ’π’™ βˆ’ 𝒙^𝟐 )+γ€–π’”π’Šπ’γ€—^(βˆ’πŸ)⁑〖((𝒙 + 𝟐)/πŸ‘)+π‘ͺγ€—

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.