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Misc 38 - Prove that 2 tan3 x dx = 1 - log2 - CBSE - Miscellaneous

Misc 38 - Chapter 7 Class 12 Integrals - Part 2
Misc 38 - Chapter 7 Class 12 Integrals - Part 3


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Misc 38 Prove that ∫_0^(πœ‹/4)β–’γ€–2 tanγ€—^3⁑π‘₯ 𝑑π‘₯=1βˆ’log⁑2 Solving L.H.S 2∫_0^(πœ‹/4)β–’γ€– tanγ€—^3⁑π‘₯ 𝑑π‘₯ = 2∫_0^(πœ‹/4)β–’γ€– tan π‘₯ tanγ€—^2⁑π‘₯ 𝑑π‘₯ = 2∫_0^(πœ‹/4)β–’γ€– tan π‘₯ (secγ€—^2⁑〖π‘₯βˆ’1)γ€— 𝑑π‘₯ = 2∫_0^(πœ‹/4)β–’γ€– tan π‘₯ secγ€—^2⁑π‘₯ 𝑑π‘₯βˆ’ 2∫_0^(πœ‹/4)β–’tan⁑〖π‘₯ 𝑑π‘₯γ€— 𝑰_𝟏 = 2∫_𝟎^(𝝅/πŸ’)β–’γ€– 𝐭𝐚𝐧 𝒙 𝒔𝒆𝒄〗^πŸβ‘π’™ 𝒅𝒙 Let t = tan x 𝑑𝑑/𝑑π‘₯ = 〖𝑠𝑒𝑐〗^2 x dt = 〖𝑠𝑒𝑐〗^2x dx Substituting, 2∫1_0^1▒〖𝑑 𝑑𝑑〗 = 2 [𝑑^2/2]_0^1 = 2 (1/2βˆ’0) =1 𝑰_𝟐= 2∫_𝟎^(𝝅/πŸ’)▒𝒕𝒂𝒏⁑〖𝒙 𝒅𝒙〗 2∫_𝟎^(𝝅/πŸ’)▒𝒕𝒂𝒏⁑〖𝒙 𝒅𝒙〗 = 2 ["log" |saec⁑π‘₯ |]_0^(πœ‹/4) = 2 ("log" √2βˆ’0) = 2 (log 2^(1/2)) = 2 (1/2 "log " 2) = log 2 Hence, = 𝐼_1βˆ’πΌ_2 = 1 βˆ’ log 2 = 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 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.