For any integer n, the value of ∫_(-π)^π  e^(cos^2 x) sin^3 (2n+1)x dx is

(a) -1           (b) 0              (c) 1           (d) 2

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This is of the form ∫_(βˆ’π‘Ž)^π‘Žβ–’π‘“(π‘₯)𝑑π‘₯ where 𝒇(𝒙)=𝒆^(〖𝒄𝒐𝒔〗^𝟐 𝒙) γ€–π¬π’Šπ’γ€—^πŸ‘ (πŸπ’+𝟏)𝒙 And, 𝑓(βˆ’π‘₯)=𝑒^〖𝒄𝒐𝒔〗^𝟐⁑(βˆ’π’™) sin^3⁑〖[(2𝑛+1)γ€— (βˆ’π’™)] Since cos (-x) = cos x 𝑓(βˆ’π‘₯)=𝑒^(γ€–π‘π‘œπ‘ γ€—^2⁑π‘₯) γ€–s𝑖𝑛〗^3⁑[βˆ’(2𝑛+1)π‘₯] Since sin (-x) = βˆ’sin x 𝒇(βˆ’π’™)=γ€–βˆ’π’†γ€—^(〖𝒄𝒐𝒔〗^πŸβ‘π’™) γ€–π’”π’Šπ’γ€—^πŸ‘β‘(πŸπ’+𝟏)𝒙 Thus, 𝑓(βˆ’π‘₯)=βˆ’π‘“(π‘₯) ∴ ∫130_(βˆ’π…)^π…β–’β€Š 𝒆^(〖𝒄𝒐𝒔〗^𝟐 𝒙) γ€–π’”π’Šπ’γ€—^πŸ‘ (πŸπ’+𝟏)𝒙𝒅𝒙=𝟎 So, the correct answer is (b)

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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.