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Ex 7.11, 14 - Using properties, evaluate integral cos5 x dx

Ex 7.11, 14 - Chapter 7 Class 12 Integrals - Part 2
Ex 7.11, 14 - Chapter 7 Class 12 Integrals - Part 3

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Transcript

Ex 7.10, 14 By using the properties of definite integrals, evaluate the integrals : ∫_0^2πœ‹β–’cos^5⁑π‘₯ 𝑑π‘₯ ∫_0^2πœ‹β–’cos^5⁑π‘₯ 𝑑π‘₯ =∫_0^πœ‹β–’cos^5⁑π‘₯ 𝑑π‘₯+∫_0^πœ‹β–’γ€–cos^5 (2Ο€βˆ’π‘₯)γ€— 𝑑π‘₯ = ∫_0^πœ‹β–’γ€–γ€–π‘π‘œπ‘ γ€—^5 π‘₯ 𝑑π‘₯+∫_0^πœ‹β–’γ€–π‘π‘œπ‘ γ€—^5 γ€— π‘₯ = 2 ∫_0^πœ‹β–’γ€–γ€–π‘π‘œπ‘ γ€—^5 π‘₯ 𝑑π‘₯γ€— Using property: ∫_0^2π‘Žβ–’γ€–π‘“(π‘₯)𝑑π‘₯=∫_0^π‘Žβ–’γ€–π‘“(π‘₯)𝑑π‘₯+∫_0^π‘Žβ–’π‘“(2π‘Žβˆ’π‘₯)𝑑π‘₯γ€—γ€— (As cos (2Ο€ βˆ’ πœƒ) = cos πœƒ) Using property: ∫_0^2π‘Žβ–’γ€–π‘“(π‘₯)𝑑π‘₯=∫_0^π‘Žβ–’γ€–π‘“(π‘₯)𝑑π‘₯+∫_0^π‘Žβ–’π‘“(2π‘Žβˆ’π‘₯)𝑑π‘₯γ€—γ€— = 2 (∫_0^(πœ‹/2)β–’γ€–γ€–π‘π‘œπ‘ γ€—^5 π‘₯ 𝑑π‘₯+∫_0^(πœ‹/2)β–’γ€–cos⁑(πœ‹βˆ’π‘₯) γ€—γ€— 𝑑π‘₯) = 2 (∫_0^(πœ‹/2)β–’γ€–γ€–π‘π‘œπ‘ γ€—^5 π‘₯ 𝑑π‘₯+∫_0^(πœ‹/2)β–’γ€–(βˆ’ cos π‘₯)^5⁑𝑑π‘₯ γ€—γ€—) = 2 (∫_0^(πœ‹/2)β–’γ€–γ€–π‘π‘œπ‘ γ€—^5 π‘₯ 𝑑π‘₯βˆ’βˆ«_0^(πœ‹/2)β–’γ€–γ€–γ€–π‘π‘œπ‘ γ€—^5 π‘₯〗⁑𝑑π‘₯ γ€—γ€—) = 2Γ—0 = 0 (cos (πœ‹βˆ’πœƒ) = – cos πœƒ)

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.