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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