Subscribe to our Youtube Channel - https://www.youtube.com/channel/UCZBx269Tl5Os5NHlSbVX4Kg
Ex 7.3, 1 - Chapter 7 Class 12 Integrals
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
Ex 7.3
Ex 7.3, 1
You are here
Ex 7.3, 2
Ex 7.3, 3 Important
Ex 7.3, 4 Important
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
Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.