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

Transcript

Example 20 Find ∫1β–’(π‘₯ sin^(βˆ’1)⁑π‘₯)/√(1 βˆ’ π‘₯^2 ) 𝑑π‘₯ Example 20 Find ∫1β–’(π‘₯ sin^(βˆ’1)⁑π‘₯)/√(1 βˆ’ π‘₯^2 ) 𝑑π‘₯ ∫1β–’(π‘₯ sin^(βˆ’1)⁑π‘₯)/√(1 βˆ’ π‘₯^2 ) 𝑑π‘₯ Let t = 〖𝑠𝑖𝑛〗^(βˆ’1) (π‘₯) 𝑑𝑑/𝑑π‘₯=1/√(1 βˆ’ π‘₯^2 ) dt = 𝑑π‘₯/√(1 βˆ’ π‘₯^2 ) So, our equation becomes ∫1β–’(π‘₯ sin^(βˆ’1)⁑π‘₯)/√(1 βˆ’ π‘₯^2 ) 𝑑π‘₯ = ∫1β–’γ€–sin⁑〖𝑑×𝑑 〗×𝑑π‘₯/√(1 βˆ’ π‘₯^2 )γ€— = ∫1β–’γ€–sin⁑〖𝑑×𝑑 γ€— 𝑑𝑑〗 =𝑑 ∫1β–’γ€–sin⁑〖𝑑 𝑑𝑑 βˆ’ ∫1β–’(𝑑(𝑑))/𝑑𝑑〗 γ€— ∫1β–’sin⁑〖𝑑 𝑑𝑑 γ€— 𝑑𝑑 = t (βˆ’cos t) βˆ’ ∫1β–’(βˆ’cos⁑𝑑 ) 𝑑𝑑 = βˆ’ t cos t + ∫1β–’cos⁑𝑑 𝑑𝑑 = βˆ’t cos t + sin t + C 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) = t and g(x) = sin t Hence we take First function :-f(x) = t Second function :- g(x) = sin t ∫1β–’γ€–sin⁑〖𝑑×𝑑 𝑑𝑑=𝑑 ∫1β–’γ€–sin⁑〖𝑑 𝑑𝑑 βˆ’ ∫1β–’(𝑑(𝑑))/𝑑𝑑〗 γ€—γ€— γ€— ∫1β–’sin⁑〖𝑑 𝑑𝑑 γ€— 𝑑𝑑 = t (βˆ’cost) βˆ’ ∫1β–’(βˆ’cos⁑𝑑 ) 𝑑𝑑 = βˆ’ t cost + ∫1β–’cos⁑𝑑 𝑑𝑑 = βˆ’t cost + sin t + C (∫1β–’sin⁑〖π‘₯ 𝑑π‘₯=βˆ’cos⁑π‘₯ γ€— " " ) (∫1β–’cos⁑〖π‘₯ 𝑑π‘₯ = sin⁑π‘₯ γ€— " " ) t = 〖𝑠𝑖𝑛〗^(βˆ’1) (π‘₯) sin t = x sin t = x 〖𝑠𝑖𝑛〗^2 𝑑 = π‘₯^2 1 βˆ’ cos^2⁑𝑑 = π‘₯^2 γ€–π‘π‘œπ‘ γ€—^2 t = 1 βˆ’ π‘₯^2 cos t = √(1βˆ’π‘₯^2 ) Now, Hence putting the values. ∫1β–’(π‘₯ 〖𝑠𝑖𝑛〗^(βˆ’1) π‘₯)/√(1βˆ’π‘₯^2 ) 𝑑π‘₯=" βˆ’t cost + sin t + C" =π’™βˆ’βˆš(πŸβˆ’π’™^𝟐 ) γ€–π’”π’Šπ’γ€—^(βˆ’πŸ) 𝒙 +𝐂

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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.