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Misc 15 Integrate the function cos^3⁑π‘₯ 𝑒^log⁑sin⁑π‘₯ ∫1▒〖𝑒^log⁑sin⁑π‘₯ cos^3〗⁑π‘₯ = ∫1▒〖𝑠𝑖𝑛 π‘₯ cos^3〗⁑〖π‘₯ 𝑑π‘₯γ€— Let t = sin x 𝑑𝑑/𝑑π‘₯=cos⁑π‘₯ 𝑑𝑑/cos⁑π‘₯ = 𝑑π‘₯ (𝑒^logβ‘π‘Ž =π‘Ž) Putting value of t and dt in our equation ∫1▒〖𝑠𝑖𝑛 π‘₯ cos^3〗⁑〖π‘₯ 𝑑π‘₯γ€— = ∫1▒〖𝑑 γ€–π‘π‘œπ‘ γ€—^3 γ€— π‘₯ 𝑑π‘₯ = ∫1▒〖𝑑 γ€–π‘π‘œπ‘ γ€—^3 γ€— π‘₯×𝑑𝑑/cos⁑π‘₯ = ∫1▒〖𝑑 γ€–π‘π‘œπ‘ γ€—^2 γ€— π‘₯ 𝑑𝑑 = ∫1▒𝑑(1βˆ’sin^2⁑π‘₯) 𝑑𝑑 = ∫1▒〖𝑑 (1βˆ’π‘‘^2 γ€—) 𝑑𝑑 = ∫1β–’(π‘‘βˆ’π‘‘^3 ) 𝑑𝑑 = 𝑑^2/2βˆ’π‘‘^4/4+ C Putting back value of 𝑑 = sin x = ((γ€–sin⁑〖π‘₯)γ€—γ€—^2)/2βˆ’(〖𝑠𝑖𝑛〗^4 π‘₯)/4+ C =(1 βˆ’ γ€–π‘π‘œπ‘ γ€—^2 π‘₯)/2βˆ’ (1 βˆ’ γ€–π‘π‘œπ‘ γ€—^2 π‘₯)^2/4+ C = (1 βˆ’γ€– π‘π‘œπ‘ γ€—^2 π‘₯)/4βˆ’((1 + γ€–π‘π‘œπ‘ γ€—^4 βˆ’2γ€–π‘π‘œπ‘ γ€—^2 π‘₯)/4)+ C = 1/2βˆ’(γ€–π‘π‘œπ‘ γ€—^2 π‘₯)/2βˆ’1/4βˆ’(γ€–π‘π‘œπ‘ γ€—^4 π‘₯)/4+(γ€–π‘π‘œπ‘ γ€—^2 π‘₯)/2+ C = ((γ€–sin⁑〖π‘₯)γ€—γ€—^2)/2βˆ’(〖𝑠𝑖𝑛〗^4 π‘₯)/4+ C = (1 βˆ’ γ€–π‘π‘œπ‘ γ€—^2 π‘₯)/2βˆ’ (1 βˆ’ γ€–π‘π‘œπ‘ γ€—^2 π‘₯)^2/4+ C = (1 βˆ’γ€– π‘π‘œπ‘ γ€—^2 π‘₯)/2βˆ’((1 + γ€–π‘π‘œπ‘ γ€—^4 βˆ’ 2γ€–π‘π‘œπ‘ γ€—^2 π‘₯)/4)+ C = 1/2βˆ’(γ€–π‘π‘œπ‘ γ€—^2 π‘₯)/2βˆ’1/4βˆ’(γ€–π‘π‘œπ‘ γ€—^4 π‘₯)/4+(γ€–π‘π‘œπ‘ γ€—^2 π‘₯)/2+ C = 1/4βˆ’(γ€–π‘π‘œπ‘ γ€—^4 π‘₯)/4+ C = (βˆ’γ€–π’„π’π’”γ€—^πŸ’ 𝒙)/πŸ’+ π‘ͺ_𝟏 (Where 𝐢_1=1/4+𝐢) = ((γ€–sin⁑〖π‘₯)γ€—γ€—^2)/2βˆ’(〖𝑠𝑖𝑛〗^4 π‘₯)/4+ C = (1 βˆ’ γ€–π‘π‘œπ‘ γ€—^2 π‘₯)/2βˆ’ (1 βˆ’ γ€–π‘π‘œπ‘ γ€—^2 π‘₯)^2/4+ C = (1 βˆ’γ€– π‘π‘œπ‘ γ€—^2 π‘₯)/2βˆ’((1 + γ€–π‘π‘œπ‘ γ€—^4 βˆ’ 2γ€–π‘π‘œπ‘ γ€—^2 π‘₯)/4)+ C = 1/2βˆ’(γ€–π‘π‘œπ‘ γ€—^2 π‘₯)/2βˆ’1/4βˆ’(γ€–π‘π‘œπ‘ γ€—^4 π‘₯)/4+(γ€–π‘π‘œπ‘ γ€—^2 π‘₯)/2+ C = 1/4βˆ’(γ€–π‘π‘œπ‘ γ€—^4 π‘₯)/4+ C = (βˆ’γ€–π’„π’π’”γ€—^πŸ’ 𝒙)/πŸ’+ π‘ͺ_𝟏

  1. Chapter 7 Class 12 Integrals
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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 14 years. He provides courses for Maths, Science and Computer Science at Teachoo