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

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  1. Chapter 7 Class 12 Integrals
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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 2π‘₯+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, . ∫1β–’γ€–sin3 (2π‘₯+1) γ€— 𝑑π‘₯=∫1β–’(3 sin⁑(2π‘₯ + 1) βˆ’ sin⁑(6π‘₯ + 3))/4 𝑑π‘₯ ∫1β–’γ€–sin3 (2π‘₯+1) γ€— 𝑑π‘₯=1/4 ∫1β–’(3 sin⁑(2π‘₯+1)βˆ’sin⁑(6π‘₯+3) ) 𝑑π‘₯ ∫1β–’γ€–sin3 (2π‘₯+1) γ€— 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 cos⁑3(2π‘₯+1)+𝐢 =βˆ’3/8 cos⁑(2π‘₯+1)+1/24 [4 cos^3⁑(2π‘₯+1)βˆ’3 cos⁑(2π‘₯+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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