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Ex 7.11, 1 By using the properties of definite integrals, evaluate the integrals : ∫_0^(πœ‹/2)β–’γ€–cos^2⁑π‘₯ 𝑑π‘₯γ€— Let 𝐈=∫_𝟎^(𝝅/𝟐)▒〖〖𝒄𝒐𝒔〗^πŸβ‘π’™ 𝒅𝒙〗 I=∫_𝟎^(𝝅/𝟐)β–’γ€–γ€–πœπ¨π¬γ€—^𝟐⁑ (𝝅/πŸβˆ’π’™)𝒅𝒙〗 I= ∫_𝟎^((𝝅 )/𝟐)▒〖〖𝐬𝐒𝐧〗^𝟐 𝒙〗⁑𝒅𝒙 Using P4 : ∫_0^π‘Žβ–’γ€–π‘“(π‘₯)𝑑π‘₯=γ€— ∫_0^π‘Žβ–’π‘“(π‘Žβˆ’π‘₯)𝑑π‘₯ (As cos (πœ‹/2βˆ’πœƒ)=sinβ‘πœƒ) …(2) Adding (1) and (2) I+I= ∫_0^(πœ‹/2)β–’γ€–cos^2⁑π‘₯ 𝑑π‘₯γ€— + ∫_0^((πœ‹ )/2)β–’γ€–sin^2 π‘₯〗⁑𝑑π‘₯ 2I= ∫_0^((πœ‹ )/2)β–’(cos^2⁑〖π‘₯+sin^2⁑π‘₯ γ€— )⁑𝑑π‘₯ 𝟐𝐈 =∫_𝟎^((𝝅 )/𝟐)β–’γ€–πŸ .〗⁑𝒅𝒙 2I=[π‘₯]_0^(πœ‹/2) 2I =πœ‹/2βˆ’0 2I =πœ‹/2 𝐈=𝝅/πŸ’

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.