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  1. Chapter 7 Class 12 Integrals
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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€— = 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 Ex 7.6, 7(Method 2) π‘₯ sin^(βˆ’1)⁑π‘₯ ∫1β–’γ€–π‘₯ sin^(βˆ’1)⁑π‘₯ γ€— 𝑑π‘₯ ∫1β–’γ€–π‘₯ sin^(βˆ’1)⁑π‘₯ γ€— 𝑑π‘₯" "=∫1β–’(sin^(βˆ’1)⁑π‘₯ ) π‘₯ 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€— = 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€— 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 ((βˆ’1)/((1)^2 βˆ’π‘₯^2 ).𝑑π‘₯+∫1β–’(1βˆ’π‘₯^2 )^(1 βˆ’ 1/2) . 𝑑π‘₯) I1 = βˆ’ 1/2 (βˆ’sin^(βˆ’1)⁑π‘₯ +∫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) 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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CA Maninder Singh
CA Maninder Singh is a Chartered Accountant for the past 8 years. He provides courses for Practical Accounts, Taxation and Efiling at teachoo.com .
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