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Ex 7.3, 1 Find the integral of sin2 (2π‘₯ + 5) ∫1β–’γ€–π’”π’Šπ’πŸ (πŸπ’™ + πŸ“) γ€— 𝒅𝒙 =∫1β–’(1 βˆ’ γ€–π‘π‘œπ‘  2〗⁑(2π‘₯ + 5))/2 𝑑π‘₯ =1/2 ∫1β–’γ€–1βˆ’cos⁑(4π‘₯+10) γ€— 𝑑π‘₯ =1/2 [∫1β–’1 𝑑π‘₯βˆ’βˆ«1β–’cos⁑(4π‘₯+10) 𝑑π‘₯] We know that 𝐜𝐨𝐬 𝟐𝜽=πŸβˆ’πŸ γ€–π’”π’Šπ’γ€—^𝟐⁑𝜽 2 sin^2 πœƒ=1βˆ’cos⁑2πœƒ sin^2 πœƒ=1/2 [1βˆ’cos⁑2πœƒ ] Replace πœƒ by (𝟐𝐱+πŸ“) sin^2 (2π‘₯+5)=(1 βˆ’ cos⁑2(2π‘₯ + 5))/2 As ∫1β–’cos⁑(π‘Žπ‘₯+𝑏) 𝑑π‘₯=sin⁑(π‘Žπ‘₯ + 𝑏)/π‘Ž+𝐢 =1/2 [π‘₯βˆ’ sin⁑(4π‘₯ + 10)/4 +𝐢] =𝒙/𝟐 βˆ’ 𝟏/πŸ– π’”π’Šπ’β‘(πŸ’π’™+𝟏𝟎)+π‘ͺ

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Davneet Singh

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