

Ex 7.3
Ex 7.3, 2
Ex 7.3, 3 Important
Ex 7.3, 4 Important
Ex 7.3, 5 You are here
Ex 7.3, 6 Important
Ex 7.3, 7
Ex 7.3, 8
Ex 7.3, 9 Important
Ex 7.3, 10 Important
Ex 7.3, 11
Ex 7.3, 12
Ex 7.3, 13 Important
Ex 7.3, 14
Ex 7.3, 15
Ex 7.3, 16 Important
Ex 7.3, 17
Ex 7.3, 18 Important
Ex 7.3, 19 Important
Ex 7.3, 20 Important
Ex 7.3, 21
Ex 7.3, 22 Important
Ex 7.3, 23 (MCQ)
Ex 7.3, 24 (MCQ) Important
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