Ex 7.3, 5 - Chapter 7 Class 12 Integrals (Term 2)
Last updated at Dec. 20, 2019 by
Last updated at Dec. 20, 2019 by
Transcript
Ex 7.3, 5 Integrate sin^3β‘π₯ cos^3 π₯ β«1βγsin^3β‘π₯ cos^3 π₯γ ππ₯ =β«1βγπ ππβ‘π₯. sin^2β‘π₯ cos^3 π₯γ ππ₯ =β«1βγπ ππβ‘π₯ (1βcos^2β‘π₯ ) cos^3 π₯γ ππ₯ =β«1βγ(1βcos^2β‘π₯ ) cos^3 π₯γ. π ππβ‘π₯ ππ₯ Let cosβ‘π₯=π‘ Differentiating w.r.t.x βsinβ‘π₯=ππ‘/ππ₯ ππ₯=ππ‘/(βsinβ‘π₯ ) (γπ ππγ^2β‘π=1βγπππ γ^2β‘π) β¦(1) Thus, our equation becomes β«1βγsin^2β‘π₯ cos^3 π₯γ ππ₯ =β«1βγ(1βcos^2β‘π₯ ) cos^3 π₯γ. π ππβ‘π₯ ππ₯" " =β«1βγ(1βπ‘^2 ) π‘^3 γ. π ππβ‘π₯Γ1/(βsinβ‘π₯ ) ππ‘" " =β«1βγβ(1βπ‘^2 ) π‘^3 γ ππ‘" " =ββ«1βγ(π‘^3βπ‘^5 ) γ ππ‘" " =β[β«1βπ‘^3 ππ‘ββ«1βπ‘^5 ππ‘] =β[π‘^(3 + 1)/(3 + 1) β π‘^(5 + 1)/(5 + 1)]+πΆ =β[π‘^4/4 β π‘^6/6]+πΆ =γβπ‘γ^4/4 + π‘^6/6+πΆ =π‘^6/6 β π‘^4/4 +πΆ Putting back value of π‘=πππ β‘π₯ =(γπππγ^π π)/π β (γπππγ^π π)/π +πͺ
Integration using trigo identities - sin^2,cos^2 etc formulae
Integration using trigo identities - sin^2,cos^2 etc formulae
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