Ex 7.10, 10 - Chapter 7 Class 12 Integrals

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〗 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)