Check sibling questions

Misc 15 - Chapter 7 Class 12 Integrals

Last updated at May 29, 2023 by

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

Misc 15 - Chapter 7 Class 12 Integrals - Part 2
Misc 15 - Chapter 7 Class 12 Integrals - Part 3

Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class

Book a free demo

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

Ask a doubt
Davneet Singh's photo - Co-founder, Teachoo

Made by

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.