Integration Full Chapter Explained - https://you.tube/Integration-Class-12

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  1. Chapter 7 Class 12 Integrals
  2. Serial order wise

Transcript

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
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.