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, 10 (sin^(βˆ’1)⁑π‘₯ )^2 ∫1β–’(sin^(βˆ’1)⁑π‘₯ )^2 𝑑π‘₯ Let sin^(βˆ’1)⁑π‘₯=πœƒ ∴ π‘₯=sinβ‘πœƒ Differentiating both sides 𝑀.π‘Ÿ.𝑑.π‘₯ 𝑑π‘₯/π‘‘πœƒ=cosβ‘πœƒ 𝑑π‘₯=cosβ‘πœƒ.π‘‘πœƒ Thus, our equation becomes ∫1β–’(sin^(βˆ’1)⁑π‘₯ )^2 𝑑π‘₯ = ∫1β–’πœƒ^2 . cosβ‘πœƒ.π‘‘πœƒ =πœƒ^2 ∫1β–’γ€–cosβ‘πœƒ.π‘‘πœƒγ€—βˆ’βˆ«1β–’(𝑑(πœƒ^2 )/π‘‘πœƒ ∫1β–’γ€–cosβ‘πœƒ.π‘‘πœƒγ€—)π‘‘πœƒ =πœƒ^2 sinβ‘πœƒβˆ’βˆ«1β–’γ€–2πœƒ.sinβ‘πœƒ. π‘‘πœƒγ€— =πœƒ^2 sinβ‘πœƒβˆ’2∫1β–’γ€–πœ½ π’”π’Šπ’β‘πœ½. π’…πœ½γ€— We know that ∫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) = ΞΈ2 and g(x) = cos ΞΈ Solving I1 ∫1β–’γ€–πœ½ π’”π’Šπ’β‘πœ½. π’…πœ½γ€— = πœƒβˆ«1β–’γ€–sinβ‘πœƒ π‘‘πœƒγ€—βˆ’βˆ«1β–’(π‘‘πœƒ/π‘‘πœƒ ∫1β–’γ€–sinβ‘πœƒ π‘‘πœƒγ€—) π‘‘πœƒ = πœƒ(βˆ’cosβ‘πœƒ )βˆ’βˆ«1β–’γ€–1.(βˆ’cosβ‘πœƒ ) γ€— π‘‘πœƒ = βˆ’πœƒ cosβ‘πœƒ+∫1β–’cosβ‘πœƒ π‘‘πœƒ = (βˆ’πœƒ cosβ‘πœƒ+sinβ‘πœƒ )+𝐢1 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) = ΞΈ and g(x) = sin ΞΈ Putting value of I1 in our equation ∫1β–’(sin^(βˆ’1)⁑π‘₯ )^2 𝑑π‘₯ = πœƒ^2 sinβ‘πœƒβˆ’2∫1β–’γ€–πœ½ π’”π’Šπ’β‘πœ½. π’…πœ½γ€— =πœƒ^2 sinβ‘πœƒβˆ’2(βˆ’πœƒ cosβ‘πœƒ+sinβ‘πœƒ+𝐢1) =πœƒ^2 sinβ‘πœƒ+2πœƒ cosβ‘πœƒβˆ’2 sinβ‘πœƒβˆ’2𝐢1 =πœƒ^2 sinβ‘πœƒ+2πœƒβˆš(1βˆ’sin^2β‘γ€–πœƒ γ€— )βˆ’2 sinβ‘πœƒβˆ’π‘ͺ𝟏 =πœƒ^2 sinβ‘πœƒ+2πœƒβˆš(1βˆ’sin^2β‘γ€–πœƒ γ€— )βˆ’2 sinβ‘πœƒ+π‘ͺ =(〖𝐬𝐒𝐧〗^(βˆ’πŸ)⁑𝒙 )^𝟐 𝒙+𝟐(〖𝐬𝐒𝐧〗^(βˆ’πŸ)⁑𝒙 ) √(γ€–πŸβˆ’γ€—β‘γ€–π’™^𝟐 γ€— )βˆ’πŸπ’™+π‘ͺ π‘ˆπ‘ π‘–π‘›π‘” πœƒ=sin^(βˆ’1)⁑π‘₯ & sinβ‘πœƒ=π‘₯

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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.