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Integrate x sin^-1 x (x sin inverse x) - Chapter 7 Class 12

Ex 7.6, 7 - Chapter 7 Class 12 Integrals - Part 2
Ex 7.6, 7 - Chapter 7 Class 12 Integrals - Part 3 Ex 7.6, 7 - Chapter 7 Class 12 Integrals - Part 4 Ex 7.6, 7 - Chapter 7 Class 12 Integrals - Part 5 Ex 7.6, 7 - Chapter 7 Class 12 Integrals - Part 6 Ex 7.6, 7 - Chapter 7 Class 12 Integrals - Part 7 Ex 7.6, 7 - Chapter 7 Class 12 Integrals - Part 8

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Ex 7.6, 7 (Method 1) π‘₯ sin^(βˆ’1)⁑π‘₯ ∫1β–’γ€–π‘₯ 〖𝑠𝑖𝑛〗^(βˆ’1) γ€— π‘₯ 𝑑π‘₯ Let x = sinβ‘πœƒ dx = cosβ‘πœƒ π‘‘πœƒ Substituting values, we get ∫1β–’γ€–π‘₯ 〖𝑠𝑖𝑛〗^(βˆ’1) γ€— π‘₯ 𝑑π‘₯ = ∫1β–’γ€–sinβ‘πœƒ γ€–π’”π’Šπ’γ€—^(βˆ’πŸ)⁑(π’”π’Šπ’β‘πœ½ ) cosβ‘πœƒ π‘‘πœƒ" " γ€— = ∫1β–’γ€–sinβ‘πœƒ 𝜽 cosβ‘πœƒ π‘‘πœƒ" " γ€— = 1/2 ∫1β–’γ€–πœƒ γ€–(2 sinγ€—β‘πœƒ cosβ‘γ€–πœƒ)γ€— π‘‘πœƒ" " γ€— = 1/2 ∫1β–’γ€–γ€–πœƒ sin〗⁑2πœƒ π‘‘πœƒγ€— (∡sin^(βˆ’1)⁑〖(sinβ‘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 = 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] = 1/2 [πœƒ (βˆ’(cos⁑2πœƒ))/2βˆ’βˆ«1.(βˆ’γ€–(cos〗⁑2πœƒ))/2 π‘‘πœƒ] = 1/2 [(βˆ’πœƒ)/2 cos⁑2πœƒ+1/2∫cos 2πœƒ π‘‘πœƒ] = 1/2 [(βˆ’πœƒ)/2 cos⁑2πœƒ+1/2 ( sin⁑2πœƒ)/2] + C = (βˆ’πœƒ)/4 π’„π’π’”β‘πŸπœ½ + 1/8 π’”π’Šπ’β‘πŸπœ½ + C = (βˆ’πœƒ)/4 (1 βˆ’ γ€–πŸπ’”π’Šπ’γ€—^𝟐 𝜽) + 1/8 Γ— 2 π’”π’Šπ’β‘πœ½ 𝐜𝐨𝐬 𝜽 + C = (βˆ’πœƒ)/4 (1 βˆ’ γ€–2𝑠𝑖𝑛〗^2 πœƒ) + 1/4 π‘ π‘–π‘›β‘πœƒ π‘π‘œπ‘  πœƒ + C = (βˆ’πœƒ)/4 (1 βˆ’ γ€–2𝑠𝑖𝑛〗^2 πœƒ) + 1/4 π‘ π‘–π‘›β‘πœƒ √(1βˆ’sin^2 πœƒ) + C = (βˆ’sin^(βˆ’1)⁑π‘₯)/4 (1 βˆ’ 2π‘₯^2 )+π‘₯/4 √(1βˆ’π‘₯^2 )+C = ((γ€–πŸπ’™γ€—^𝟐 βˆ’ 𝟏))/πŸ’ γ€–π’”π’Šπ’γ€—^(βˆ’πŸ) 𝒙+𝒙/πŸ’ √(πŸβˆ’π’™^𝟐 )+𝐂 Using π‘₯=sinβ‘πœƒ ⟹ πœƒ=sin^(βˆ’1)⁑π‘₯ Ex 7.6, 7 (Method 2) π‘₯ sin^(βˆ’1)⁑π‘₯ ∫1β–’γ€–π‘₯ sin^(βˆ’1)⁑π‘₯ γ€— 𝑑π‘₯ =sin^(βˆ’1)⁑π‘₯ ∫1β–’π‘₯ 𝑑π‘₯βˆ’βˆ«1β–’(𝑑(sin^(βˆ’1)⁑π‘₯ )/𝑑π‘₯ ∫1β–’γ€–π‘₯ .𝑑π‘₯γ€—)𝑑π‘₯ = sin^(βˆ’1)⁑π‘₯. π‘₯^2/2 βˆ’βˆ«1β–’γ€–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) = sin–1 x = π‘₯^2/2 sin^(βˆ’1)⁑π‘₯βˆ’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€— ("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‘₯ sin^(βˆ’1)⁑π‘₯ 𝑑π‘₯γ€—=π‘₯^2/2 sin^(βˆ’1)⁑π‘₯βˆ’(sin^(βˆ’1)⁑π‘₯/4 βˆ’ π‘₯/4 √(1βˆ’π‘₯^2 )β‘βˆ’ 𝐢1/2) =π‘₯^2/2 sin^(βˆ’1)⁑π‘₯βˆ’sin^(βˆ’1)⁑π‘₯/4+ π‘₯/4 √(1βˆ’π‘₯^2 )⁑+ 𝐢1/2 = sin^(βˆ’1)⁑π‘₯ (π‘₯^2/2βˆ’1/4)+ π‘₯/4 √(1βˆ’π‘₯^2 )⁑+ 𝐢 = sin^(βˆ’1)⁑π‘₯ ((γ€–2π‘₯γ€—^2 βˆ’1)/4)+ π‘₯/4 √(1βˆ’π‘₯^2 )⁑+ 𝐢 = ((γ€–πŸπ’™γ€—^𝟐 βˆ’πŸ)/πŸ’) γ€–π’”π’Šπ’γ€—^(βˆ’πŸ)⁑𝒙+ 𝒙/πŸ’ √(πŸβˆ’π’™^𝟐 )⁑+ π‘ͺ

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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.