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Ex 7.6, 22 - Integrate sin^-1 (2x / 1 + x^2) - Teachoo - Ex 7.6

Ex 7.6, 22 - Chapter 7 Class 12 Integrals - Part 2
Ex 7.6, 22 - Chapter 7 Class 12 Integrals - Part 3
Ex 7.6, 22 - Chapter 7 Class 12 Integrals - Part 4
Ex 7.6, 22 - Chapter 7 Class 12 Integrals - Part 5

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Ex 7.6, 22 Integrate the function sin^(βˆ’1) (2π‘₯/(1 + π‘₯2)) Simplifying the given function sin^(βˆ’1) (2π‘₯/(1 + π‘₯2)) Let π‘₯=tan⁑𝑑 ∴ 𝑑=tan^(βˆ’1)⁑(π‘₯) ∴ sin^(βˆ’1) (2π‘₯/(1 + π‘₯2))=sin^(βˆ’1) ((2 tan⁑𝑑" " )/(1 + tan^2⁑𝑑 )) =sin^(βˆ’1) (sin⁑2𝑑 ) = 2t Ex 7.6, 22 Integrate the function sin^(βˆ’1) (2π‘₯/(1 + π‘₯2)) Simplifying the given function sin^(βˆ’1) (2π‘₯/(1 + π‘₯2)) Let π‘₯=tan⁑𝑑 ∴ 𝑑=tan^(βˆ’1)⁑(π‘₯) ∴ sin^(βˆ’1) (2π‘₯/(1 + π‘₯2))=sin^(βˆ’1) ((2 tan⁑𝑑" " )/(1 + tan^2⁑𝑑 )) =sin^(βˆ’1) (sin⁑2𝑑 ) = 2t (π‘ˆπ‘ π‘–π‘›π‘” sin⁑2πœƒ=(2 tanβ‘πœƒ" " )/(1 + tan^2β‘πœƒ )) (π‘ˆπ‘ π‘–π‘›π‘” sin^(βˆ’1) (sin⁑π‘₯ )=π‘₯) =2 tan^(βˆ’1)⁑π‘₯ Thus, our function becomes ∫1β–’γ€–sin^(βˆ’1) (2π‘₯/(1 + π‘₯2)) 𝑑π‘₯γ€— = 2∫1β–’γ€–tan^(βˆ’1)⁑π‘₯ 𝑑π‘₯γ€— =2∫1β–’γ€–(tan^(βˆ’1) π‘₯) 1.𝑑π‘₯ " " γ€— = 2 tan^(βˆ’1) π‘₯∫1β–’γ€–1 .γ€— 𝑑π‘₯βˆ’2∫1β–’(𝑑(tan^(βˆ’1)⁑π‘₯ )/𝑑π‘₯ ∫1β–’γ€–1 .𝑑π‘₯γ€—) 𝑑π‘₯ = 2tan^(βˆ’1) π‘₯ (π‘₯)βˆ’2∫1β–’1/(1 + π‘₯^2 ) . π‘₯ . 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 (π‘ˆπ‘ π‘–π‘›π‘” 𝑑=tan^(βˆ’1)⁑(π‘₯) ) = 2π‘₯ tan^(βˆ’1) π‘₯βˆ’2∫1β–’π‘₯/(1 + π‘₯^2 ) . 𝑑π‘₯ 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€–sin^(βˆ’1) (2π‘₯/(1 + π‘₯2)) γ€— .𝑑π‘₯=2∫1β–’γ€–" " tan^(βˆ’1) π‘₯" " γ€— .𝑑π‘₯ =2π‘₯ tan^(βˆ’1) π‘₯βˆ’2∫1β–’π‘₯/(1 + π‘₯^2 ) . 𝑑π‘₯ =2π‘₯ tan^(βˆ’1) π‘₯βˆ’2(1/2 γ€–log 〗⁑|1+π‘₯^2 |+𝐢1) (As t = 1 + x2) =2π‘₯ tan^(βˆ’1) π‘₯βˆ’γ€–log 〗⁑|1+π‘₯^2 |βˆ’2𝐢1 =πŸπ’™ 〖𝒕𝒂𝒏〗^(βˆ’πŸ) π’™βˆ’γ€–π’π’π’ˆ 〗⁑(𝟏+𝒙^𝟐 )+π‘ͺ As 1 + x2 is always positive |1+π‘₯^2 | = 1 + x2

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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.