Example 10 (ii) - Chapter 7 Class 12 Integrals

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Example 10 Find the following integrals: (ii) ∫1▒(𝑥 + 3)/√(5 − 4𝑥 − 𝑥^2 ) 𝑑𝑥 We can write numerator as 𝑥+3= A𝑑/𝑑𝑥 [−𝑥^2−4𝑥+5]+ B 𝑥+3= A[−2𝑥−4]+ B 𝑥+3=−2𝐴𝑥−4A+B Finding A & B Comparing coefficient of 𝑥 𝑥=−2𝐴𝑥 1 =−2A A=(−1)/( 2) Comparing constant term 3=−4A+B 3=−4((−1)/( 2))+B 3=2+B B=3−2=1 Now, we know that 𝑥+3=A[−2𝑥−4]+B 𝑥+3=(−1)/( 2) [−2𝑥−4]+1 Now, our equation is ∫1▒(𝑥+3)/√(5 − 4𝑥 + 𝑥^2 ).𝑑𝑥=∫1▒(1/2 [−2𝑥 − 4]+1)/√(5 − 4𝑥 − 𝑥^2 ).𝑑𝑥 =∫1▒((−1)/2 [−2𝑥 − 4]+1)/√(5 − 4𝑥 + 𝑥^2 ) 𝑑𝑥+∫1▒1/√(5 − 4𝑥 − 𝑥^2 ) 𝑑𝑥 =(−1)/2 ∫1▒(−2𝑥 − 4)/√(5 − 4𝑥 + 𝑥^2 ) 𝑑𝑥+∫1▒1/√(5 − 4𝑥 − 𝑥^2 ) 𝑑𝑥 Solving I1 I1 = (−1)/2 ∫1▒〖(−2𝑥 − 4)/√(−𝑥^2 − 4𝑥 + 5).𝑑𝑥〗 Let t = −𝑥^2 − 4𝑥 + 5 Differentiating both sides w.r.t. 𝑥 −2𝑥 − 4 =𝑑𝑡/𝑑𝑥 𝑑𝑥=𝑑𝑡/(− 2𝑥 − 4) Now, I1 = (−1)/( 2) ∫1▒〖(−2𝑥 − 4)/√(−𝑥^2 − 4𝑥 + 5).𝑑𝑥〗 Putting the values of (−𝑥^2−4𝑥+5) and 𝑑𝑥 I1 = (−1)/( 2) ∫1▒〖(−2𝑥 − 4)/√𝑡.𝑑𝑡/((−2𝑥 −4) )〗 I1 = (−1)/( 2) ∫1▒〖1/√𝑡.𝑑𝑡〗 I1 = (−1)/( 2) ∫1▒〖1/(𝑡)^(1/2) .𝑑𝑡〗 I1 = (−1)/( 2) ∫1▒〖(𝑡)^((−1)/2) 𝑑𝑡〗 I1 = (−1)/( 2) 𝑡^((−1)/( 2) + 1)/(((−1)/( 2) + 1) )+𝐶1 I1 = (−1)/( 2) 𝑡^(1/( 2) )/((1/2) )+𝐶1 I1 = − 𝑡^(1/( 2) )+𝐶1 I1 = −√𝑡+𝐶1 Putting back 𝑡=𝑥^2−4𝑥+5) I1 = −√(−𝑥^2−4𝑥+5)+𝐶1 Solving 𝑰𝟐 I2 = ∫1▒〖1/√(5 − 4𝑥 − 𝑥^2 ).𝑑𝑥〗 I2 = ∫1▒〖1/√(−(𝑥^2 + 4𝑥 − 5) ) 𝑑𝑥〗 I2 = ∫1▒〖1/√(−[𝑥^2 + 2(2)(𝑥) − 5] ) 𝑑𝑥〗 I2 = ∫1▒〖1/√(−[𝑥^2 + 2(2)(𝑥) + (2)^2 − (2)^2 − 5] ) 𝑑𝑥〗 I2 = ∫1▒〖1/√(−[(𝑥 + 2)^2 − 4 − 5] ) 𝑑𝑥〗 I2 = ∫1▒〖1/√(−[(𝑥 + 2)^2 − 9] ) 𝑑𝑥〗 I2 = ∫1▒〖1/√(9 − (𝑥 + 2)^2 ) 𝑑𝑥〗 I2 = ∫1▒〖1/√((3)^2 − (𝑥 + 2)^2 ) 𝑑𝑥〗 It is of form ∫1▒〖1/√(𝑎^2 − 𝑥^2 ).𝑑𝑥=〖𝑠𝑖𝑛〗^(−1)⁡〖𝑥/𝑎+𝐶2〗 〗 Replacing 𝑥 by (𝑥+2) and a by 3, we get I2 = sin^(−1)⁡〖((𝑥 + 2)/3)+𝐶2〗 Now, Putting values of I1 and I2 in (1) ∫1▒〖(𝑥 + 3)/√(5 − 4𝑥 − 𝑥^2 ).𝑑𝑥=(−1)/( 2) ∫1▒(− 2𝑥 − 4)/√(5 − 4𝑥 − 𝑥^2 )〗+∫1▒1/√(5 − 4𝑥 − 𝑥^2 ) 𝑑𝑥 =−√(5 − 4𝑥 − 𝑥^2 )+𝐶1+sin^(−1)⁡((𝑥 + 2)/3)+𝐶2 =−√(𝟓 − 𝟒𝒙 − 𝒙^𝟐 )+〖𝒔𝒊𝒏〗^(−𝟏)⁡〖((𝒙 + 𝟐)/𝟑)+𝑪〗