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Misc 35 Prove that ∫_0^(πœ‹/2)β–’sin^3⁑π‘₯ 𝑑π‘₯=2/3 Solving L.H.S ∫_0^(πœ‹/2)β–’sin^3⁑π‘₯ 𝑑π‘₯ = ∫_0^(πœ‹/2)β–’γ€– γ€–sin π‘₯ (sinγ€—^2⁑〖π‘₯)γ€— γ€— 𝑑π‘₯ = ∫_0^(πœ‹/2)▒𝑠𝑖𝑛⁑〖π‘₯ (1βˆ’γ€–π‘π‘œπ‘ γ€—^2 π‘₯)γ€— 𝑑π‘₯ = ∫_0^(πœ‹/2)▒𝑠𝑖𝑛⁑〖π‘₯ 𝑑π‘₯βˆ’ ∫1_0^(πœ‹/2)β–’γ€–sin⁑〖π‘₯ γ€–π‘π‘œπ‘ γ€—^2 π‘₯γ€— 𝑑π‘₯γ€—γ€— 𝑰_𝟏 ∫1_0^(πœ‹/2)β–’sin⁑〖π‘₯ 𝑑π‘₯γ€— = βˆ’ [cos⁑π‘₯ ]_0^(πœ‹/2) = βˆ’[0βˆ’1] = 1 𝑰_𝟐 ∫1_0^(πœ‹/2)β–’sin⁑〖π‘₯ γ€–π‘π‘œπ‘ γ€—^2 π‘₯ 𝑑π‘₯γ€— Let t = cos x 𝑑𝑑/𝑑π‘₯ = - sin x dt = βˆ’ sin x dx Substituting, ∫1_0^1β–’sin⁑〖π‘₯×𝑑^2Γ—γ€— 𝑑𝑑/(βˆ’sin⁑〖π‘₯ γ€— ) = βˆ’βˆ«1_1^0▒〖𝑑^2 𝑑𝑑〗 = γ€–βˆ’[𝑑^3/3]γ€—_1^0 = βˆ’("0 βˆ’ " 1/3)=βˆ’((βˆ’1)/3) = 1/3 L.H.S = 𝐼_1βˆ’ 𝐼_2 = 1 βˆ’ 1/3 = 𝟐/πŸ‘ = R.H.S Hence, proved.

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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 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.