Ex 7.6, 9 - Chapter 7 Class 12 Integrals (Term 2)

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Ex 7.6, 9 (Method 1) 𝑥 〖c𝑜𝑠^(−1)〗⁡𝑥 ∫1▒𝑥 cos^(−1)⁡〖𝑥 𝑑𝑥〗 Let x = cos⁡𝜃 dx = − sin⁡〖𝜃 𝑑𝜃〗 Substituting values, we get ∫1▒𝑥 cos^(−1)⁡〖𝑥 𝑑𝑥 〗 = ∫1▒cos⁡𝜃 〖𝒄𝒐𝒔〗^(−𝟏) (𝒄𝒐𝒔⁡𝜽)(−sin⁡〖𝜃 )𝑑𝜃〗 = −∫1▒〖〖 sin〗⁡〖𝜃 〗 𝜽 cos⁡𝜃 𝑑𝜃〗 = (−1)/2 ∫1▒〖𝜃 (2sin〗⁡〖𝜃 cos⁡〖𝜃) 〗 𝑑𝜃〗 = (−1)/2 ∫1▒〖〖𝜃 sin〗⁡2𝜃 𝑑𝜃〗 Ex 7.6, 9 (Method 1) 𝑥 〖c𝑜𝑠^(−1)〗⁡𝑥 ∫1▒𝑥 cos^(−1)⁡〖𝑥 𝑑𝑥〗 Let x = cos⁡𝜃 dx = − sin⁡〖𝜃 𝑑𝜃〗 Substituting values, we get ∫1▒𝑥 cos^(−1)⁡〖𝑥 𝑑𝑥 〗 = ∫1▒cos⁡𝜃 〖𝒄𝒐𝒔〗^(−𝟏) (𝒄𝒐𝒔⁡𝜽)(−sin⁡〖𝜃 )𝑑𝜃〗 = −∫1▒〖〖 sin〗⁡〖𝜃 〗 𝜽 cos⁡𝜃 𝑑𝜃〗 = (−1)/2 ∫1▒〖𝜃 (2sin〗⁡〖𝜃 cos⁡〖𝜃) 〗 𝑑𝜃〗 = (−1)/2 ∫1▒〖〖𝜃 sin〗⁡2𝜃 𝑑𝜃〗 (∵〖𝑐𝑜𝑠〗^(−1)⁡〖(𝑐𝑜𝑠⁡𝜃)〗 = 𝜃) (∵2⁡sin⁡𝑥 cos⁡𝑥 = ⁡sin⁡2𝑥 ) Integrating by parts ∫1▒〖𝑓(𝑥) 𝑔⁡(𝑥) 〗 𝑑𝑥=𝑓(𝑥) ∫1▒𝑔(𝑥) 𝑑𝑥−∫1▒(𝑓^′ (𝑥) ∫1▒𝑔(𝑥) 𝑑𝑥) 𝑑𝑥 Putting f(x) = θ and g(x) = sin 2θ = (−1)/2 [𝜃∫sin⁡2𝜃 𝑑𝜃−∫(𝑑𝜃/𝑑𝜃∫sin 2𝜃 𝑑𝜃)𝑑𝜃] = (−1)/2 [𝜃 (−(cos⁡2𝜃))/2−∫1.(−〖(cos〗⁡2𝜃))/2 𝑑𝜃] = (−1)/2 [(−𝜃)/2 cos⁡2𝜃+1/2∫cos 2𝜃 𝑑𝜃] = (−1)/2 [(−𝜃)/2 cos⁡2𝜃+1/2 ( sin⁡2𝜃)/2] + C = 𝜃/4 𝒄𝒐𝒔⁡𝟐𝜽 – 1/8 𝒔𝒊𝒏⁡𝟐𝜽 + C = 𝜃/4 (〖𝟐𝒄𝒐𝒔〗^𝟐 𝜽 – 1) – 1/8 × 2 𝒔𝒊𝒏⁡𝜽 𝐜𝐨𝐬 𝜽 + C = 𝜃/4 (〖2𝑐𝑜𝑠〗^2 𝜃 – 1) – 1/4 𝑠𝑖𝑛⁡𝜃 𝑐𝑜𝑠 𝜃 + C = 𝜃/4 (〖2𝑐𝑜𝑠〗^2 𝜃 – 1) – 1/4 𝑐𝑜𝑠⁡𝜃 √(1−cos^2 𝜃) + C = cos^(−1)⁡𝑥/4 (2𝑥^2−1)−𝑥/4 √(1−𝑥^2 )+C = ((〖𝟐𝒙〗^𝟐 − 𝟏))/𝟒 〖𝒄𝒐𝒔〗^(−𝟏) 𝒙−𝒙/𝟒 √(𝟏−𝒙^𝟐 )+𝐂 Using 𝑥=cos⁡𝜃 ⟹ 𝜃=cos^(−1)⁡𝑥 Ex 7.6, 9 (Method 2) 𝑥 cos^(−1)⁡𝑥 ∫1▒〖𝑥 cos^(−1)⁡𝑥 〗 𝑑𝑥 =cos^(−1)⁡𝑥 ∫1▒𝑥 𝑑𝑥−∫1▒(𝑑(cos^(−1)⁡𝑥 )/𝑑𝑥 ∫1▒〖𝑥 .𝑑𝑥〗)𝑑𝑥 = cos^(−1)⁡𝑥. 𝑥^2/2 −∫1▒〖(−1)/√(1 − 𝑥^2 ) . 𝑥^2/2. 𝑑𝑥〗 Now we know that ∫1▒〖𝑓(𝑥) 𝑔⁡(𝑥) 〗 𝑑𝑥=𝑓(𝑥) ∫1▒𝑔(𝑥) 𝑑𝑥−∫1▒(𝑓′(𝑥)∫1▒𝑔(𝑥) 𝑑𝑥) 𝑑𝑥 Putting f(x) = x and g(x) = cos–1 x = 𝑥^2/2 cos^(−1)⁡𝑥+1/2 ∫1▒〖𝑥^2/√(1 − 𝑥^2 ) 𝑑𝑥〗 Solving I1 I1 = 1/2 ∫1▒〖𝑥^2/√(1 − 𝑥^2 ) 𝑑𝑥〗 Multiplying and dividing by −1. I1 = 1/2 ∫1▒〖〖−𝑥〗^2/( −√(1 − 𝑥^2 )) 𝑑𝑥〗 I1 = (−1)/2 ∫1▒〖〖−𝑥〗^2/( √(1 − 𝑥^2 )) 𝑑𝑥〗 I1 =(−1)/2 ∫1▒〖〖−1 + 1 − 𝑥〗^2/( √(1 − 𝑥^2 )) 𝑑𝑥〗 …(1) ("Adding and subtracting" "1 in numerator" ) I1 = (−1)/2 ∫1▒〖((−1)/√(1 − 𝑥^2 )+(1 −〖 𝑥〗^2)/√(1 − 𝑥^2 )) 𝑑𝑥〗 I1 = (−1)/2 (∫1▒〖(−𝟏)/√((𝟏)^𝟐 −𝒙^𝟐 ).𝒅𝒙〗+∫1▒(1−𝑥^2 )^(1 − 1/2) . 𝑑𝑥) I1 = (−1)/2 (−〖𝒔𝒊𝒏〗^(−𝟏)⁡𝒙 +∫1▒(1−𝑥^2 )^(1/2) . 𝑑𝑥) I1 = (−1)/2 (−sin^(−1)⁡𝑥 +∫1▒√((1)^2−𝑥^2 ). 𝑑𝑥) I1 = (−1)/2 (−sin^(−1)⁡𝑥 +𝑥/2 √(1−𝑥^2 )+(1)^2/2 sin^(−1)⁡〖 𝑥/1〗+𝐶1) Using ∫1▒〖𝑑𝑥/√(𝑎^2 − 𝑥^2 )=sin^(−1)⁡〖𝑥/𝑎〗 +𝐶〗 𝑈𝑠𝑖𝑛𝑔 √(𝑎^2−𝑥^2 ) 𝑑𝑥=1/𝑥 𝑥√(𝑎^2−𝑥^2 )+𝑎^2/2 sin^(−1)⁡〖𝑥/𝑎〗+𝐶 I1 = (−1)/2 (−sin^(−1)⁡𝑥+1/2 〖 sin〗^(−1)⁡𝑥+𝑥/2 √(1−𝑥^2 )⁡+ 𝐶1) I1 = (−1)/2 ((−sin^(−1)⁡𝑥)/2 +𝑥/2 √(1−𝑥^2 )⁡+ 𝐶1) I1 = sin^(−1)⁡𝑥/4 − 𝑥/4 √(1−𝑥^2 )⁡− 𝐶1/2 Putting the value of I1 in eq. (1), we get ∫1▒〖𝑥 cos^(−1)⁡𝑥 𝑑𝑥〗=𝑥^2/2 cos^(−1)⁡𝑥+(sin^(−1)⁡𝑥/4 − 𝑥/4 √(1−𝑥^2 )⁡− 𝐶1/2) =𝑥^2/2 cos^(−1)⁡𝑥+1/4 sin^(−1)⁡𝑥− 𝑥/4 √(1−𝑥^2 )⁡+ 𝐶1/2 =𝑥^2/2 cos^(−1)⁡𝑥+1/4 (𝜋/2−cos^(−1)⁡𝑥 )− 𝑥/4 √(1−𝑥^2 )⁡+ 𝐶1/2 =𝑥^2/2 cos^(−1)⁡𝑥+𝜋/8−1/4 cos^(−1)⁡𝑥− 𝑥/4 √(1−𝑥^2 )⁡+ 𝐶1/2 =𝑥^2/2 cos^(−1)⁡𝑥−1/4 cos^(−1)⁡𝑥− 𝑥/4 √(1−𝑥^2 )⁡+ 𝐶1/2 +𝜋/8 =𝑥^2/2 cos^(−1)⁡𝑥−1/4 cos^(−1)⁡𝑥− 𝑥/4 √(1−𝑥^2 )⁡+ 𝐶 = cos^(−1)⁡𝑥 (𝑥^2/2−1/4)− 𝑥/4 √(1−𝑥^2 )⁡+ 𝐶 = cos^(−1)⁡𝑥 ((〖2𝑥〗^2 − 1)/4)− 𝑥/4 √(1−𝑥^2 )⁡+ 𝐶 = ((〖𝟐𝒙〗^𝟐+ 𝟏)/𝟒) 〖𝒄𝒐𝒔〗^(−𝟏)⁡𝒙− 𝒙/𝟒 √(𝟏−𝒙^𝟐 )⁡+ 𝑪