Integration Full Chapter Explained - Integration Class 12 - Everything you need
Misc 15 - Chapter 7 Class 12 Integrals
Last updated at Dec. 23, 2019 by Teachoo
Transcript
Misc 15 Integrate the function cos^3β‘π₯ π^logβ‘sinβ‘π₯ β«1βγπ^logβ‘sinβ‘π₯ cos^3γβ‘π₯ = β«1βγπ ππ π₯ cos^3γβ‘γπ₯ ππ₯γ Let t = sin x ππ‘/ππ₯=cosβ‘π₯ ππ‘/cosβ‘π₯ = ππ₯ (π^logβ‘π =π) Putting value of t and dt in our equation β«1βγπ ππ π₯ cos^3γβ‘γπ₯ ππ₯γ = β«1βγπ‘ γπππ γ^3 γ π₯ ππ₯ = β«1βγπ‘ γπππ γ^3 γ π₯Γππ‘/cosβ‘π₯ = β«1βγπ‘ γπππ γ^2 γ π₯ ππ‘ = β«1βπ‘(1βsin^2β‘π₯) ππ‘ = β«1βγπ‘ (1βπ‘^2 γ) ππ‘ = β«1β(π‘βπ‘^3 ) ππ‘ = π‘^2/2βπ‘^4/4+ C Putting back value of π‘ = sin x = ((γsinβ‘γπ₯)γγ^2)/2β(γπ ππγ^4 π₯)/4+ C =(1 β γπππ γ^2 π₯)/2β (1 β γπππ γ^2 π₯)^2/4+ C = (1 βγ πππ γ^2 π₯)/4β((1 + γπππ γ^4 β2γπππ γ^2 π₯)/4)+ C = 1/2β(γπππ γ^2 π₯)/2β1/4β(γπππ γ^4 π₯)/4+(γπππ γ^2 π₯)/2+ C = ((γsinβ‘γπ₯)γγ^2)/2β(γπ ππγ^4 π₯)/4+ C = (1 β γπππ γ^2 π₯)/2β (1 β γπππ γ^2 π₯)^2/4+ C = (1 βγ πππ γ^2 π₯)/2β((1 + γπππ γ^4 β 2γπππ γ^2 π₯)/4)+ C = 1/2β(γπππ γ^2 π₯)/2β1/4β(γπππ γ^4 π₯)/4+(γπππ γ^2 π₯)/2+ C = 1/4β(γπππ γ^4 π₯)/4+ C = (βγπππγ^π π)/π+ πͺ_π (Where πΆ_1=1/4+πΆ) = ((γsinβ‘γπ₯)γγ^2)/2β(γπ ππγ^4 π₯)/4+ C = (1 β γπππ γ^2 π₯)/2β (1 β γπππ γ^2 π₯)^2/4+ C = (1 βγ πππ γ^2 π₯)/2β((1 + γπππ γ^4 β 2γπππ γ^2 π₯)/4)+ C = 1/2β(γπππ γ^2 π₯)/2β1/4β(γπππ γ^4 π₯)/4+(γπππ γ^2 π₯)/2+ C = 1/4β(γπππ γ^4 π₯)/4+ C = (βγπππγ^π π)/π+ πͺ_π
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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.