# Ex 7.3, 5 - Class 12

Last updated at May 29, 2018 by Teachoo

Last updated at May 29, 2018 by Teachoo

Transcript

Ex 7.3, 5 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₯ ) Putting πππ π₯=π‘ & ππ₯=ππ‘/(βπ ππβ‘π₯ ) in (1) 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 =β π‘^4/4 + π‘^6/6+πΆ =π‘^6/6 β π‘^4/4 +πΆ Putting values of π‘=πππ β‘π₯ =(γπππγ^π π)/π β (γπππγ^π π)/π +πͺ

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About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.