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Chapter 7 Class 12 Integrals
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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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