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

Transcript

Ex 7.6, 13 Integrate the function - tan^(βˆ’1) π‘₯ ∫1β–’γ€–" " tan^(βˆ’1) π‘₯" " γ€— .𝑑π‘₯=∫1β–’γ€–(tan^(βˆ’1) π‘₯) 1.𝑑π‘₯ " " γ€— = tan^(βˆ’1) π‘₯∫1β–’γ€–1 .γ€— 𝑑π‘₯βˆ’βˆ«1β–’(𝑑(tan^(βˆ’1)⁑π‘₯ )/𝑑π‘₯ ∫1β–’γ€–1 .𝑑π‘₯γ€—) 𝑑π‘₯ = tan^(βˆ’1) π‘₯ (π‘₯)βˆ’βˆ«1β–’1/(1 + π‘₯^2 ) . π‘₯ . 𝑑π‘₯ = π‘₯ tan^(βˆ’1) π‘₯βˆ’βˆ«1β–’π‘₯/(1 + π‘₯^2 ) . 𝑑π‘₯ 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) = tan–1 x and g(x) = 1 Solving I1 I1 = ∫1β–’π‘₯/(1 + π‘₯^2 ) . 𝑑π‘₯" " Let 1 + π‘₯^2=𝑑 Differentiating both sides 𝑀.π‘Ÿ.𝑑.π‘₯ 0 + 2π‘₯=𝑑𝑑/𝑑π‘₯ 𝑑π‘₯=𝑑𝑑/2π‘₯ Our equation becomes I1 = ∫1β–’π‘₯/(1 + π‘₯^2 ) . 𝑑π‘₯" " Putting the value of (1+π‘₯^2 ) = t and 𝑑π‘₯ = 𝑑𝑑/( 2π‘₯) , we get I1 = ∫1β–’π‘₯/𝑑 . 𝑑𝑑/2π‘₯ I1 = 1/2 ∫1β–’1/𝑑 . 𝑑𝑑 I1 = 1/2 log⁑〖 |𝑑|γ€—+𝐢1 I1 = 1/2 log⁑〖 |1+π‘₯^2 |γ€—+𝐢1 Putting the value of I1 in (1) , ∫1β–’γ€–" " tan^(βˆ’1) π‘₯" " γ€— .𝑑π‘₯=π‘₯ tan^(βˆ’1) π‘₯βˆ’βˆ«1β–’π‘₯/(1 + π‘₯^2 ) . 𝑑π‘₯ =π‘₯ tan^(βˆ’1) π‘₯βˆ’(1/2 γ€–log 〗⁑|1+π‘₯^2 |+𝐢1) =π‘₯ tan^(βˆ’1) π‘₯βˆ’1/2 γ€–log 〗⁑|1+π‘₯^2 |βˆ’πΆ1 =𝒙 〖𝒕𝒂𝒏〗^(βˆ’πŸ) π’™βˆ’πŸ/𝟐 γ€–π’π’π’ˆ 〗⁑(𝟏+𝒙^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.