1. Chapter 7 Class 12 Integrals (Term 2)
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Misc 15 - Chapter 7 Class 12 Integrals (Term 2)

Last updated at Dec. 23, 2019 by

Misc 15 - Integrate cos3 x elog sinx - Chapter 7 NCERT - Miscellaneous

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Misc 15 - Chapter 7 Class 12 Integrals - Part 2

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Misc 15 - Chapter 7 Class 12 Integrals - Part 3

  1. Chapter 7 Class 12 Integrals (Term 2)
  2. Serial order wise

    3. Miscellaneous

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Transcript

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

Chapter 7 Class 12 Integrals (Term 2)
Serial order wise

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.