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Ex 7.6, 9 - Integrate x cos-1 x - Chapter 7 Class 12 - Ex 7.6

Ex 7.6, 9 - Chapter 7 Class 12 Integrals - Part 2
Ex 7.6, 9 - Chapter 7 Class 12 Integrals - Part 3 Ex 7.6, 9 - Chapter 7 Class 12 Integrals - Part 4 Ex 7.6, 9 - Chapter 7 Class 12 Integrals - Part 5 Ex 7.6, 9 - Chapter 7 Class 12 Integrals - Part 6 Ex 7.6, 9 - Chapter 7 Class 12 Integrals - Part 7 Ex 7.6, 9 - Chapter 7 Class 12 Integrals - Part 8


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Ex 7.6, 9 (Method 1) π‘₯ γ€–cπ‘œπ‘ ^(βˆ’1)〗⁑π‘₯ ∫1β–’π‘₯ cos^(βˆ’1)⁑〖π‘₯ 𝑑π‘₯γ€— Let x = cosβ‘πœƒ dx = βˆ’ sinβ‘γ€–πœƒ π‘‘πœƒγ€— Substituting values, we get ∫1β–’π‘₯ cos^(βˆ’1)⁑〖π‘₯ 𝑑π‘₯ γ€— = ∫1β–’cosβ‘πœƒ 〖𝒄𝒐𝒔〗^(βˆ’πŸ) (π’„π’π’”β‘πœ½)(βˆ’sinβ‘γ€–πœƒ )π‘‘πœƒγ€— = βˆ’βˆ«1β–’γ€–γ€– sinγ€—β‘γ€–πœƒ γ€— 𝜽 cosβ‘πœƒ π‘‘πœƒγ€— = (βˆ’1)/2 ∫1β–’γ€–πœƒ (2sinγ€—β‘γ€–πœƒ cosβ‘γ€–πœƒ) γ€— π‘‘πœƒγ€— = (βˆ’1)/2 ∫1β–’γ€–γ€–πœƒ sin〗⁑2πœƒ π‘‘πœƒγ€— Ex 7.6, 9 (Method 1) π‘₯ γ€–cπ‘œπ‘ ^(βˆ’1)〗⁑π‘₯ ∫1β–’π‘₯ cos^(βˆ’1)⁑〖π‘₯ 𝑑π‘₯γ€— Let x = cosβ‘πœƒ dx = βˆ’ sinβ‘γ€–πœƒ π‘‘πœƒγ€— Substituting values, we get ∫1β–’π‘₯ cos^(βˆ’1)⁑〖π‘₯ 𝑑π‘₯ γ€— = ∫1β–’cosβ‘πœƒ 〖𝒄𝒐𝒔〗^(βˆ’πŸ) (π’„π’π’”β‘πœ½)(βˆ’sinβ‘γ€–πœƒ )π‘‘πœƒγ€— = βˆ’βˆ«1β–’γ€–γ€– sinγ€—β‘γ€–πœƒ γ€— 𝜽 cosβ‘πœƒ π‘‘πœƒγ€— = (βˆ’1)/2 ∫1β–’γ€–πœƒ (2sinγ€—β‘γ€–πœƒ cosβ‘γ€–πœƒ) γ€— π‘‘πœƒγ€— = (βˆ’1)/2 ∫1β–’γ€–γ€–πœƒ sin〗⁑2πœƒ π‘‘πœƒγ€— (βˆ΅γ€–π‘π‘œπ‘ γ€—^(βˆ’1)⁑〖(π‘π‘œπ‘ β‘πœƒ)γ€— = πœƒ) (∡2⁑sin⁑π‘₯ cos⁑π‘₯ = ⁑sin⁑2π‘₯ ) Integrating by parts ∫1▒〖𝑓(π‘₯) 𝑔⁑(π‘₯) γ€— 𝑑π‘₯=𝑓(π‘₯) ∫1▒𝑔(π‘₯) 𝑑π‘₯βˆ’βˆ«1β–’(𝑓^β€² (π‘₯) ∫1▒𝑔(π‘₯) 𝑑π‘₯) 𝑑π‘₯ Putting f(x) = ΞΈ and g(x) = sin 2ΞΈ = (βˆ’1)/2 [πœƒβˆ«sin⁑2πœƒ π‘‘πœƒβˆ’βˆ«(π‘‘πœƒ/π‘‘πœƒβˆ«sin 2πœƒ π‘‘πœƒ)π‘‘πœƒ] = (βˆ’1)/2 [πœƒ (βˆ’(cos⁑2πœƒ))/2βˆ’βˆ«1.(βˆ’γ€–(cos〗⁑2πœƒ))/2 π‘‘πœƒ] = (βˆ’1)/2 [(βˆ’πœƒ)/2 cos⁑2πœƒ+1/2∫cos 2πœƒ π‘‘πœƒ] = (βˆ’1)/2 [(βˆ’πœƒ)/2 cos⁑2πœƒ+1/2 ( sin⁑2πœƒ)/2] + C = πœƒ/4 π’„π’π’”β‘πŸπœ½ – 1/8 π’”π’Šπ’β‘πŸπœ½ + C = πœƒ/4 (γ€–πŸπ’„π’π’”γ€—^𝟐 𝜽 – 1) – 1/8 Γ— 2 π’”π’Šπ’β‘πœ½ 𝐜𝐨𝐬 𝜽 + C = πœƒ/4 (γ€–2π‘π‘œπ‘ γ€—^2 πœƒ – 1) – 1/4 π‘ π‘–π‘›β‘πœƒ π‘π‘œπ‘  πœƒ + C = πœƒ/4 (γ€–2π‘π‘œπ‘ γ€—^2 πœƒ – 1) – 1/4 π‘π‘œπ‘ β‘πœƒ √(1βˆ’cos^2 πœƒ) + C = cos^(βˆ’1)⁑π‘₯/4 (2π‘₯^2βˆ’1)βˆ’π‘₯/4 √(1βˆ’π‘₯^2 )+C = ((γ€–πŸπ’™γ€—^𝟐 βˆ’ 𝟏))/πŸ’ 〖𝒄𝒐𝒔〗^(βˆ’πŸ) π’™βˆ’π’™/πŸ’ √(πŸβˆ’π’™^𝟐 )+𝐂 Using π‘₯=cosβ‘πœƒ ⟹ πœƒ=cos^(βˆ’1)⁑π‘₯ Ex 7.6, 9 (Method 2) π‘₯ cos^(βˆ’1)⁑π‘₯ ∫1β–’γ€–π‘₯ cos^(βˆ’1)⁑π‘₯ γ€— 𝑑π‘₯ =cos^(βˆ’1)⁑π‘₯ ∫1β–’π‘₯ 𝑑π‘₯βˆ’βˆ«1β–’(𝑑(cos^(βˆ’1)⁑π‘₯ )/𝑑π‘₯ ∫1β–’γ€–π‘₯ .𝑑π‘₯γ€—)𝑑π‘₯ = cos^(βˆ’1)⁑π‘₯. π‘₯^2/2 βˆ’βˆ«1β–’γ€–(βˆ’1)/√(1 βˆ’ π‘₯^2 ) . π‘₯^2/2. 𝑑π‘₯γ€— Now we know that ∫1▒〖𝑓(π‘₯) 𝑔⁑(π‘₯) γ€— 𝑑π‘₯=𝑓(π‘₯) ∫1▒𝑔(π‘₯) 𝑑π‘₯βˆ’βˆ«1β–’(𝑓′(π‘₯)∫1▒𝑔(π‘₯) 𝑑π‘₯) 𝑑π‘₯ Putting f(x) = x and g(x) = cos–1 x = π‘₯^2/2 cos^(βˆ’1)⁑π‘₯+1/2 ∫1β–’γ€–π‘₯^2/√(1 βˆ’ π‘₯^2 ) 𝑑π‘₯γ€— Solving I1 I1 = 1/2 ∫1β–’γ€–π‘₯^2/√(1 βˆ’ π‘₯^2 ) 𝑑π‘₯γ€— Multiplying and dividing by βˆ’1. I1 = 1/2 ∫1β–’γ€–γ€–βˆ’π‘₯γ€—^2/( βˆ’βˆš(1 βˆ’ π‘₯^2 )) 𝑑π‘₯γ€— I1 = (βˆ’1)/2 ∫1β–’γ€–γ€–βˆ’π‘₯γ€—^2/( √(1 βˆ’ π‘₯^2 )) 𝑑π‘₯γ€— I1 =(βˆ’1)/2 ∫1β–’γ€–γ€–βˆ’1 + 1 βˆ’ π‘₯γ€—^2/( √(1 βˆ’ π‘₯^2 )) 𝑑π‘₯γ€— …(1) ("Adding and subtracting" "1 in numerator" ) I1 = (βˆ’1)/2 ∫1β–’γ€–((βˆ’1)/√(1 βˆ’ π‘₯^2 )+(1 βˆ’γ€– π‘₯γ€—^2)/√(1 βˆ’ π‘₯^2 )) 𝑑π‘₯γ€— I1 = (βˆ’1)/2 (∫1β–’γ€–(βˆ’πŸ)/√((𝟏)^𝟐 βˆ’π’™^𝟐 ).𝒅𝒙〗+∫1β–’(1βˆ’π‘₯^2 )^(1 βˆ’ 1/2) . 𝑑π‘₯) I1 = (βˆ’1)/2 (βˆ’γ€–π’”π’Šπ’γ€—^(βˆ’πŸ)⁑𝒙 +∫1β–’(1βˆ’π‘₯^2 )^(1/2) . 𝑑π‘₯) I1 = (βˆ’1)/2 (βˆ’sin^(βˆ’1)⁑π‘₯ +∫1β–’βˆš((1)^2βˆ’π‘₯^2 ). 𝑑π‘₯) I1 = (βˆ’1)/2 (βˆ’sin^(βˆ’1)⁑π‘₯ +π‘₯/2 √(1βˆ’π‘₯^2 )+(1)^2/2 sin^(βˆ’1)⁑〖 π‘₯/1γ€—+𝐢1) Using ∫1▒〖𝑑π‘₯/√(π‘Ž^2 βˆ’ π‘₯^2 )=sin^(βˆ’1)⁑〖π‘₯/π‘Žγ€— +𝐢〗 π‘ˆπ‘ π‘–π‘›π‘” √(π‘Ž^2βˆ’π‘₯^2 ) 𝑑π‘₯=1/π‘₯ π‘₯√(π‘Ž^2βˆ’π‘₯^2 )+π‘Ž^2/2 sin^(βˆ’1)⁑〖π‘₯/π‘Žγ€—+𝐢 I1 = (βˆ’1)/2 (βˆ’sin^(βˆ’1)⁑π‘₯+1/2 γ€– sinγ€—^(βˆ’1)⁑π‘₯+π‘₯/2 √(1βˆ’π‘₯^2 )⁑+ 𝐢1) I1 = (βˆ’1)/2 ((βˆ’sin^(βˆ’1)⁑π‘₯)/2 +π‘₯/2 √(1βˆ’π‘₯^2 )⁑+ 𝐢1) I1 = sin^(βˆ’1)⁑π‘₯/4 βˆ’ π‘₯/4 √(1βˆ’π‘₯^2 )β‘βˆ’ 𝐢1/2 Putting the value of I1 in eq. (1), we get ∫1β–’γ€–π‘₯ cos^(βˆ’1)⁑π‘₯ 𝑑π‘₯γ€—=π‘₯^2/2 cos^(βˆ’1)⁑π‘₯+(sin^(βˆ’1)⁑π‘₯/4 βˆ’ π‘₯/4 √(1βˆ’π‘₯^2 )β‘βˆ’ 𝐢1/2) =π‘₯^2/2 cos^(βˆ’1)⁑π‘₯+1/4 sin^(βˆ’1)⁑π‘₯βˆ’ π‘₯/4 √(1βˆ’π‘₯^2 )⁑+ 𝐢1/2 =π‘₯^2/2 cos^(βˆ’1)⁑π‘₯+1/4 (πœ‹/2βˆ’cos^(βˆ’1)⁑π‘₯ )βˆ’ π‘₯/4 √(1βˆ’π‘₯^2 )⁑+ 𝐢1/2 =π‘₯^2/2 cos^(βˆ’1)⁑π‘₯+πœ‹/8βˆ’1/4 cos^(βˆ’1)⁑π‘₯βˆ’ π‘₯/4 √(1βˆ’π‘₯^2 )⁑+ 𝐢1/2 =π‘₯^2/2 cos^(βˆ’1)⁑π‘₯βˆ’1/4 cos^(βˆ’1)⁑π‘₯βˆ’ π‘₯/4 √(1βˆ’π‘₯^2 )⁑+ 𝐢1/2 +πœ‹/8 =π‘₯^2/2 cos^(βˆ’1)⁑π‘₯βˆ’1/4 cos^(βˆ’1)⁑π‘₯βˆ’ π‘₯/4 √(1βˆ’π‘₯^2 )⁑+ 𝐢 = cos^(βˆ’1)⁑π‘₯ (π‘₯^2/2βˆ’1/4)βˆ’ π‘₯/4 √(1βˆ’π‘₯^2 )⁑+ 𝐢 = cos^(βˆ’1)⁑π‘₯ ((γ€–2π‘₯γ€—^2 βˆ’ 1)/4)βˆ’ π‘₯/4 √(1βˆ’π‘₯^2 )⁑+ 𝐢 = ((γ€–πŸπ’™γ€—^𝟐+ 𝟏)/πŸ’) 〖𝒄𝒐𝒔〗^(βˆ’πŸ)β‘π’™βˆ’ 𝒙/πŸ’ √(πŸβˆ’π’™^𝟐 )⁑+ π‘ͺ

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.