Ex 7.3, 1 - Chapter 7 Class 12 Integrals (Term 2)
Last updated at Dec. 20, 2019 by Teachoo
Last updated at Dec. 20, 2019 by Teachoo
Transcript
Ex 7.3, 1 Find the integral of sin2 (2π₯ + 5) β«1βγsin2 (2π₯ + 5) γ ππ₯ =β«1β(1 β γπππ 2γβ‘(2π₯ + 5))/2 ππ₯ =1/2 β«1βγ1βcosβ‘(4π₯+10) γ ππ₯ =1/2 [β«1β1 ππ₯ββ«1βcosβ‘(4π₯+10) ππ₯] We know that cos 2π=1β2 sin^2β‘π 2 sin^2 π=1βcosβ‘2π sin^2 π=1/2 [1βcosβ‘2π ] Replace π by (2π₯+5) sin^2 (2π₯+5)=(1 β cosβ‘2(2π₯ + 5))/2 (As β«1βcosβ‘(ππ₯+π) ππ₯=sinβ‘(ππ₯ + π)/π+πΆ) =1/2 [π₯β sinβ‘(4π₯ + 10)/4 +πΆ] =π/π β π/π πππβ‘(ππ+ππ)+πͺ
Integration using trigo identities - 2x formulae
Integration using trigo identities - 2x formulae
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