This is of the form ∫_(−𝑎)^𝑎▒𝑓(𝑥)𝑑𝑥 where 𝒇(𝒙)=𝒆^(〖𝒄𝒐𝒔〗^𝟐 𝒙) 〖𝐬𝒊𝒏〗^𝟑 (𝟐𝒏+𝟏)𝒙 And, 𝑓(−𝑥)=𝑒^〖𝒄𝒐𝒔〗^𝟐⁡(−𝒙) sin^3⁡〖[(2𝑛+1)〗 (−𝒙)] Since cos (-x) = cos x 𝑓(−𝑥)=𝑒^(〖𝑐𝑜𝑠〗^2⁡𝑥) 〖s𝑖𝑛〗^3⁡[−(2𝑛+1)𝑥] Since sin (-x) = −sin x 𝒇(−𝒙)=〖−𝒆〗^(〖𝒄𝒐𝒔〗^𝟐⁡𝒙) 〖𝒔𝒊𝒏〗^𝟑⁡(𝟐𝒏+𝟏)𝒙 Thus, 𝑓(−𝑥)=−𝑓(𝑥) ∴ ∫130_(−𝝅)^𝝅▒ 𝒆^(〖𝒄𝒐𝒔〗^𝟐 𝒙) 〖𝒔𝒊𝒏〗^𝟑 (𝟐𝒏+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.

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