Slide10.JPG

Slide11.JPG
Slide12.JPG

  1. Chapter 7 Class 12 Integrals
  2. Serial order wise

Transcript

Ex 7.3, 4 sin3 (2π‘₯ + 1) We know that sin⁑3πœƒ=3 sinβ‘πœƒβˆ’4 sin^3β‘πœƒ 4 sin^3β‘πœƒ=3 sinβ‘πœƒβˆ’sin⁑3πœƒ sin^3β‘πœƒ=(3 sinβ‘πœƒ βˆ’ sin⁑3πœƒ)/4 Replace πœƒ by πŸπ’™+𝟏 sin^3⁑(2π‘₯+1)=(3 sin⁑(2π‘₯ + 1) βˆ’ sin⁑3(2π‘₯ + 1))/4 sin^3⁑(2π‘₯+1)=(3 sin⁑(2π‘₯ + 1) βˆ’ sin⁑(6π‘₯ + 3))/4 Thus, our equation becomes . ∫1β–’γ€–sin3 (2π‘₯+1) γ€— 𝑑π‘₯ =∫1β–’(3 sin⁑(2π‘₯ + 1) βˆ’ sin⁑(6π‘₯ + 3))/4 𝑑π‘₯ =1/4 ∫1β–’(3 sin⁑(2π‘₯+1)βˆ’sin⁑(6π‘₯+3) ) 𝑑π‘₯ =1/4 [3∫1β–’sin⁑(2π‘₯+1) 𝑑π‘₯βˆ’βˆ«1β–’sin⁑(6π‘₯+3) 𝑑π‘₯] =1/4 [3•×1/2 (βˆ’cos⁑(2π‘₯+1) )βˆ’1/6 (βˆ’cos⁑(6π‘₯+3)+𝐢)" " ] =1/4 [(βˆ’3)/2 cos⁑(2π‘₯+1)+1/6 cos⁑(6π‘₯+3) ]+𝐢 =(βˆ’3)/8 cos⁑(2π‘₯+1)+1/24 π’„π’π’”β‘πŸ‘(πŸπ’™+𝟏)+𝐢 ∫1β–’sin⁑(π‘Žπ‘₯+𝑏) 𝑑π‘₯=βˆ’γ€–π‘π‘œπ‘  〗⁑(π‘Žπ‘₯ + 𝑏)/π‘Ž +𝐢 We know that π‘π‘œπ‘ β‘3πœƒ=4 γ€–π‘π‘œπ‘ γ€—^3β‘πœƒβˆ’3 π‘π‘œπ‘ β‘πœƒ Replace πœƒ by 2π‘₯+1 π‘π‘œπ‘ β‘3(2π‘₯+1)=4 γ€–π‘π‘œπ‘ γ€—^3⁑(2π‘₯+1)βˆ’3 π‘π‘œπ‘ β‘(2π‘₯+1) =(βˆ’3)/8 cos⁑(2π‘₯+1)+1/24 [πŸ’ 〖𝒄𝒐𝒔〗^πŸ‘β‘(πŸπ’™+𝟏)βˆ’πŸ‘ 𝒄𝒐𝒔⁑(πŸπ’™+𝟏) ]+𝐢 =(βˆ’3)/8 cos⁑(2π‘₯+1)+4/24 cos^3⁑(2π‘₯+1)βˆ’3/24 cos⁑(2π‘₯+1)+𝐢 =(βˆ’3)/8 cos⁑(2π‘₯+1)+1/6 cos^3⁑(2π‘₯+1)βˆ’1/8 cos⁑(2π‘₯+1)+𝐢 =(βˆ’3)/8 cos⁑(2π‘₯+1)βˆ’1/8 cos⁑(2π‘₯+1)+1/6 cos^3⁑(2π‘₯+1)+𝐢 =(βˆ’4)/8 cos⁑(2π‘₯+1)+1/6 cos^3⁑(2π‘₯+1)+𝐢 =(βˆ’πŸ)/𝟐 𝒄𝒐𝒔⁑(πŸπ’™+𝟏)+𝟏/πŸ” 〖𝒄𝒐𝒔〗^πŸ‘β‘(πŸπ’™+𝟏)+π‘ͺ

About the Author

Davneet Singh's photo - Teacher, Computer Engineer, Marketer
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.