Β  Misc 35 - Prove that definite integral sin3 x dx = 2/3 - Miscellaneous - Miscellaneous

part 2 - Misc 35 - Miscellaneous - Serial order wise - Chapter 7 Class 12 Integrals
part 3 - Misc 35 - Miscellaneous - Serial order wise - Chapter 7 Class 12 Integrals

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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 and Computer Science at Teachoo