Ex 7.3, 1 - Chapter 7 Class 12 Integrals
Last updated at April 16, 2024 by Teachoo
Ex 7.3
Ex 7.3, 1
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Transcript
Ex 7.3, 1 Find the integral of sin2 (2π₯ + 5) β«1βγππππ (ππ + π) γ π π =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 ππ¨π¬ ππ½=πβπ γπππγ^πβ‘π½ 2 sin^2 π=1βcosβ‘2π sin^2 π=1/2 [1βcosβ‘2π ] Replace π by (ππ±+π) 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
