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Ex 7.11, 10 - Evaluate (2 log sin x - log sin 2x) dx - Ex 7.11

Ex 7.11, 10 - Chapter 7 Class 12 Integrals - Part 2
Ex 7.11, 10 - Chapter 7 Class 12 Integrals - Part 3

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Transcript

Ex 7.10, 10 By using the properties of definite integrals, evaluate the integrals : ∫_0^(𝜋/2)▒〖 (2 log⁡sin⁡𝑥 −log⁡sin⁡2𝑥 ) 〗 𝑑𝑥 Let I1=∫_0^(𝜋/2)▒〖 (2 log⁡sin⁡𝑥 −log⁡sin⁡2𝑥 ) 〗 𝑑𝑥 I1= ∫_0^(𝜋/2)▒〖 [2 log⁡sin⁡𝑥 −𝑙𝑜𝑔(2 sin⁡〖𝑥 cos⁡𝑥 〗 )] 〗 𝑑𝑥 I1= ∫_0^(𝜋/2)▒〖 [2 log⁡sin⁡𝑥 −log⁡2−log⁡sin⁡〖𝑥−log⁡cos⁡𝑥 〗 ] 〗 𝑑𝑥 I1= ∫_0^(𝜋/2)▒〖 [log⁡sin⁡𝑥 −𝑙𝑜𝑔2−log⁡cos⁡𝑥 ] 〗 𝑑𝑥 I1= ∫_0^(𝜋/2)▒log⁡sin⁡〖𝑥 𝑑𝑥〗 −∫_0^(𝜋/2)▒〖log⁡2𝑑𝑥−∫_0^(𝜋/2)▒log⁡cos⁡〖𝑥 𝑑𝑥〗 〗 Solving I2 I2=∫_0^(𝜋/2)▒log⁡cos⁡〖𝑥 𝑑𝑥〗 ∴ I2= ∫_0^(𝜋/2)▒log⁡𝑐𝑜𝑠(𝜋/2−𝑥)𝑑𝑥 I2=∫_0^(𝜋/2)▒log⁡sin⁡〖𝑥 𝑑𝑥〗 Put the value of I2 in (1) i.e. I1 ∴ I1= ∫_0^(𝜋/2)▒log⁡sin⁡〖𝑥 𝑑𝑥〗 −∫_0^(𝜋/2)▒〖log 2〗⁡𝑑𝑥 −∫_𝟎^(𝝅/𝟐)▒𝐥𝐨𝐠⁡𝐜𝐨𝐬⁡〖𝒙 𝒅𝒙〗 I1= ∫_0^(𝜋/2)▒log⁡sin⁡〖𝑥 𝑑𝑥〗 −∫_0^(𝜋/2)▒〖log 2〗⁡𝑑𝑥 −∫_𝟎^(𝝅/𝟐)▒𝒍𝒐𝒈⁡𝐬𝐢𝐧⁡〖𝒙 𝒅𝒙〗 Using the property P4 : ∫_0^𝑎▒〖𝑓(𝑥)𝑑𝑥=〗 ∫_0^𝑎▒𝑓(𝑎−𝑥)𝑑𝑥 I1= – ∫_0^(𝜋/2)▒〖log 2〗⁡𝑑𝑥 I1= – log 2∫_0^(𝜋/2)▒𝑑𝑥 I1= – log 2[𝑥]_0^(𝜋/2) I1= – log 2[𝜋/2−0] I1= – log 2×𝜋/2 I1= log⁡〖〖 (2)〗^(−1) 〗 [𝜋/2] I1= 𝝅/𝟐 𝐥𝐨𝐠 (𝟏/𝟐) (𝐴𝑠 log⁡〖2 𝑖𝑠 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡〗 ) (█("Using the property " @𝑎 log⁡𝑏 = log⁡𝑏𝑎))

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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 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.