Integration using trigo identities - 3x formulae

Chapter 7 Class 12 Integrals
Concept wise

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Ex 7.3, 4 sin3 (2π₯ + 1) We know that sinβ‘3π=3 sinβ‘πβ4 sin^3β‘π 4 sin^3β‘π=3 sinβ‘πβsinβ‘3π sin^3β‘π=(3 sinβ‘π β sinβ‘3π)/4 Replace π by ππ+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«1βγsin3 (2π₯+1) γ ππ₯ =β«1β(3 sinβ‘(2π₯ + 1) β sinβ‘(6π₯ + 3))/4 ππ₯ =1/4 β«1β(3 sinβ‘(2π₯+1)βsinβ‘(6π₯+3) ) ππ₯ =1/4 [3β«1βsinβ‘(2π₯+1) ππ₯ββ«1βsinβ‘(6π₯+3) ππ₯] =1/4 [3β€ΆΓ1/2 (βcosβ‘(2π₯+1) )β1/6 (βcosβ‘(6π₯+3)+πΆ)" " ] =1/4 [(β3)/2 cosβ‘(2π₯+1)+1/6 cosβ‘(6π₯+3) ]+πΆ =(β3)/8 cosβ‘(2π₯+1)+1/24 πππβ‘π(ππ+π)+πΆ β«1βsinβ‘(ππ₯+π) ππ₯=βγπππ  γβ‘(ππ₯ + π)/π +πΆ We know that πππ β‘3π=4 γπππ γ^3β‘πβ3 πππ β‘π Replace π by 2π₯+1 πππ β‘3(2π₯+1)=4 γπππ γ^3β‘(2π₯+1)β3 πππ β‘(2π₯+1) =(β3)/8 cosβ‘(2π₯+1)+1/24 [π γπππγ^πβ‘(ππ+π)βπ πππβ‘(ππ+π) ]+πΆ =(β3)/8 cosβ‘(2π₯+1)+4/24 cos^3β‘(2π₯+1)β3/24 cosβ‘(2π₯+1)+πΆ =(β3)/8 cosβ‘(2π₯+1)+1/6 cos^3β‘(2π₯+1)β1/8 cosβ‘(2π₯+1)+πΆ =(β3)/8 cosβ‘(2π₯+1)β1/8 cosβ‘(2π₯+1)+1/6 cos^3β‘(2π₯+1)+πΆ =(β4)/8 cosβ‘(2π₯+1)+1/6 cos^3β‘(2π₯+1)+πΆ =(βπ)/π πππβ‘(ππ+π)+π/π γπππγ^πβ‘(ππ+π)+πͺ