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Integration of tan inverse x - Ex 7.6, 13 - Chapter 7 Class 12

Ex 7.6, 13 - Chapter 7 Class 12 Integrals - Part 2
Ex 7.6, 13 - Chapter 7 Class 12 Integrals - Part 3


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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 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.