Last updated at Feb. 25, 2025 by Teachoo
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
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About the Author
Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 15 years. He provides courses for Maths, Science and Computer Science at Teachoo