Misc 36 - Chapter 7 Class 12 Integrals

Last updated at April 16, 2024 by Teachoo

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Misc 36 Prove that ∫_0^(𝜋/4)▒〖2 tan〗^3⁡𝑥 𝑑𝑥=1−log⁡2 Solving L.H.S 2∫_0^(𝜋/4)▒〖 tan〗^3⁡𝑥 𝑑𝑥 = 2∫_0^(𝜋/4)▒〖 tan 𝑥 tan〗^2⁡𝑥 𝑑𝑥 = 2∫_0^(𝜋/4)▒〖 tan 𝑥 (sec〗^2⁡〖𝑥−1)〗 𝑑𝑥 = 2∫_0^(𝜋/4)▒〖 tan 𝑥 sec〗^2⁡𝑥 𝑑𝑥− 2∫_0^(𝜋/4)▒tan⁡〖𝑥 𝑑𝑥〗 𝑰_𝟏 = 2∫_𝟎^(𝝅/𝟒)▒〖 𝐭𝐚𝐧 𝒙 𝒔𝒆𝒄〗^𝟐⁡𝒙 𝒅𝒙 Let t = tan x 𝑑𝑡/𝑑𝑥 = 〖𝑠𝑒𝑐〗^2 x dt = 〖𝑠𝑒𝑐〗^2x dx Substituting, 2∫1_0^1▒〖𝑡 𝑑𝑡〗 = 2 [𝑡^2/2]_0^1 = 2 (1/2−0) =1 𝑰_𝟐= 2∫_𝟎^(𝝅/𝟒)▒𝒕𝒂𝒏⁡〖𝒙 𝒅𝒙〗 2∫_𝟎^(𝝅/𝟒)▒𝒕𝒂𝒏⁡〖𝒙 𝒅𝒙〗 = 2 ["log" |sec⁡𝑥 |]_0^(𝜋/4) = 2 ("log" √2−0) = 2 (log 2^(1/2)) = 2 (1/2 "log " 2) = log 2 Hence, = 𝐼_1−𝐼_2 = 1 − log 2 = R.H.S Hence, proved.

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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

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