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Ex 7.10
Ex 7.10, 2
Ex 7.10, 3 Important
Ex 7.10, 4
Ex 7.10, 5 Important
Ex 7.10, 6
Ex 7.10,7 Important
Ex 7.10,8 Important
Ex 7.10, 9
Ex 7.10, 10 Important You are here
Ex 7.10, 11 Important
Ex 7.10, 12 Important
Ex 7.10, 13
Ex 7.10, 14
Ex 7.10, 15
Ex 7.10, 16 Important
Ex 7.10, 17
Ex 7.10, 18 Important
Ex 7.10, 19
Ex 7.10, 20 (MCQ) Important
Ex 7.10, 21 (MCQ) Important
Last updated at May 29, 2023 by Teachoo
Ex 7.10, 10 By using the properties of definite integrals, evaluate the integrals : ∫_0^(𝜋/2)▒〖 (2 logsin𝑥 −logsin2𝑥 ) 〗 𝑑𝑥 Let I1=∫_0^(𝜋/2)▒〖 (2 logsin𝑥 −logsin2𝑥 ) 〗 𝑑𝑥 I1= ∫_0^(𝜋/2)▒〖 [2 logsin𝑥 −𝑙𝑜𝑔(2 sin〖𝑥 cos𝑥 〗 )] 〗 𝑑𝑥 I1= ∫_0^(𝜋/2)▒〖 [2 logsin𝑥 −log2−logsin〖𝑥−logcos𝑥 〗 ] 〗 𝑑𝑥 I1= ∫_0^(𝜋/2)▒〖 [logsin𝑥 −𝑙𝑜𝑔2−logcos𝑥 ] 〗 𝑑𝑥 I1= ∫_0^(𝜋/2)▒logsin〖𝑥 𝑑𝑥〗 −∫_0^(𝜋/2)▒〖log2𝑑𝑥−∫_0^(𝜋/2)▒logcos〖𝑥 𝑑𝑥〗 〗 Solving I2 I2=∫_0^(𝜋/2)▒logcos〖𝑥 𝑑𝑥〗 ∴ I2= ∫_0^(𝜋/2)▒log𝑐𝑜𝑠(𝜋/2−𝑥)𝑑𝑥 I2=∫_0^(𝜋/2)▒logsin〖𝑥 𝑑𝑥〗 Put the value of I2 in (1) i.e. I1 ∴ I1= ∫_0^(𝜋/2)▒logsin〖𝑥 𝑑𝑥〗 −∫_0^(𝜋/2)▒〖log 2〗𝑑𝑥 −∫_𝟎^(𝝅/𝟐)▒𝐥𝐨𝐠𝐜𝐨𝐬〖𝒙 𝒅𝒙〗 I1= ∫_0^(𝜋/2)▒logsin〖𝑥 𝑑𝑥〗 −∫_0^(𝜋/2)▒〖log 2〗𝑑𝑥 −∫_𝟎^(𝝅/𝟐)▒𝒍𝒐𝒈𝐬𝐢𝐧〖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 log𝑏 = log𝑏𝑎))