Integration Full Chapter Explained - Integration Class 12 - Everything you need


Last updated at Dec. 20, 2019 by Teachoo
Transcript
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
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
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
Ex 7.3, 24 Important
About the Author