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Ex 7.3, 4 - Integrate sin3 (2x + 1) - Chapter 7 Class 12

Ex 7.3, 4 - Chapter 7 Class 12 Integrals - Part 2
Ex 7.3, 4 - Chapter 7 Class 12 Integrals - Part 3


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

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.