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Ex 7.3
Ex 7.3, 2
Ex 7.3, 3 Important
Ex 7.3, 4 Important You are here
Ex 7.3, 5
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
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Ex 7.3, 16 Important
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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
Last updated at May 29, 2023 by Teachoo
Ex 7.3, 4 sin3 (2π₯ + 1) We know that sinβ‘3π=3 sinβ‘πβ4 sin^3β‘π 4 sin^3β‘π=3 sinβ‘πβsinβ‘3π sin^3β‘π=(3 sinβ‘π β sinβ‘3π)/4 Replace π by ππ+π sin^3β‘(2π₯+1)=(3 sinβ‘(2π₯ + 1) β sinβ‘3(2π₯ + 1))/4 sin^3β‘(2π₯+1)=(3 sinβ‘(2π₯ + 1) β sinβ‘(6π₯ + 3))/4 Thus, our equation becomes . β«1βγsin3 (2π₯+1) γ ππ₯ =β«1β(3 sinβ‘(2π₯ + 1) β sinβ‘(6π₯ + 3))/4 ππ₯ =1/4 β«1β(3 sinβ‘(2π₯+1)βsinβ‘(6π₯+3) ) ππ₯ =1/4 [3β«1βsinβ‘(2π₯+1) ππ₯ββ«1βsinβ‘(6π₯+3) ππ₯] =1/4 [3•Γ1/2 (βcosβ‘(2π₯+1) )β1/6 (βcosβ‘(6π₯+3)+πΆ)" " ] =1/4 [(β3)/2 cosβ‘(2π₯+1)+1/6 cosβ‘(6π₯+3) ]+πΆ =(β3)/8 cosβ‘(2π₯+1)+1/24 πππβ‘π(ππ+π)+πΆ β«1βsinβ‘(ππ₯+π) ππ₯=βγπππ γβ‘(ππ₯ + π)/π +πΆ We know that πππ β‘3π=4 γπππ γ^3β‘πβ3 πππ β‘π Replace π by 2π₯+1 πππ β‘3(2π₯+1)=4 γπππ γ^3β‘(2π₯+1)β3 πππ β‘(2π₯+1) =(β3)/8 cosβ‘(2π₯+1)+1/24 [π γπππγ^πβ‘(ππ+π)βπ πππβ‘(ππ+π) ]+πΆ =(β3)/8 cosβ‘(2π₯+1)+4/24 cos^3β‘(2π₯+1)β3/24 cosβ‘(2π₯+1)+πΆ =(β3)/8 cosβ‘(2π₯+1)+1/6 cos^3β‘(2π₯+1)β1/8 cosβ‘(2π₯+1)+πΆ =(β3)/8 cosβ‘(2π₯+1)β1/8 cosβ‘(2π₯+1)+1/6 cos^3β‘(2π₯+1)+πΆ =(β4)/8 cosβ‘(2π₯+1)+1/6 cos^3β‘(2π₯+1)+πΆ =(βπ)/π πππβ‘(ππ+π)+π/π γπππγ^πβ‘(ππ+π)+πͺ