Ex 7.10, 16 - Chapter 7 Class 12 Integrals

Transcript

Ex 7.10, 16 By using the properties of definite integrals, evaluate the integrals : ∫_0^𝜋▒log⁡(1+cos⁡𝑥 ) 𝑑𝑥 Let I=∫_𝟎^𝝅▒𝒍𝒐𝒈⁡(𝟏+𝒄𝒐𝒔⁡𝒙 ) 𝒅𝒙 ∴ I=∫_0^𝜋▒log⁡(1+𝑐𝑜𝑠(𝜋−𝑥)) 𝑑𝑥 𝐈=∫_𝟎^𝝅▒𝐥𝐨𝐠⁡(𝟏−𝒄𝒐𝒔⁡𝒙 ) 𝑑𝑥 Adding (1) and (2) i.e. (1) + (2) I+I=∫_0^𝜋▒𝑙𝑜𝑔(1+cos⁡𝑥 )𝑑𝑥+∫_0^𝜋▒𝑙𝑜𝑔(1−cos⁡𝑥 )𝑑𝑥 𝟐𝐈=∫_𝟎^𝝅▒[𝒍𝒐𝒈(𝟏+𝒄𝒐𝒔⁡𝒙 )+𝒍𝒐𝒈(𝟏−𝒄𝒐𝒔⁡𝒙 )]𝒅𝒙 "Using log(a) + log(b)" = "log(a.b)") 2I=∫_0^𝜋▒𝒍𝒐𝒈[(𝟏+𝒄𝒐𝒔⁡𝒙 )(𝟏−𝒄𝒐𝒔⁡𝒙 )]𝒅𝒙 2I=∫_0^𝜋▒𝑙𝑜𝑔[1−cos^2⁡𝑥 ]𝑑𝑥 2I=∫_0^𝜋▒𝑙𝑜𝑔(〖𝑠𝑖𝑛〗^2⁡𝑥 )𝑑𝑥 Using log 𝒂^𝒃=𝒃 𝒍𝒐𝒈⁡𝒂 2I=∫_0^𝜋▒〖2 𝑙𝑜𝑔(𝑠𝑖𝑛⁡𝑥 )𝑑𝑥〗 2I=2∫_0^𝜋▒𝑙𝑜𝑔(𝑠𝑖𝑛⁡𝑥 )𝑑𝑥 𝐈=∫_𝟎^𝝅▒𝒍𝒐𝒈(𝒔𝒊𝒏⁡𝒙 )𝒅𝒙 Here, 𝒇(𝒙)=log⁡sin⁡𝑥 f(𝝅−𝒙)=𝑙𝑜𝑔[𝑠𝑖𝑛(𝜋−𝑥)]𝑑𝑥 =𝑙𝑜𝑔(sin⁡𝑥 )𝑑𝑥 =𝒇(𝒙) Therefore, I=∫_𝟎^𝝅▒𝒍𝒐𝒈(𝐬𝐢𝐧⁡𝒙 )𝒅𝒙=𝟐∫_𝟎^(𝝅/𝟐)▒𝒍𝒐𝒈(𝒔𝒊𝒏⁡𝒙 )𝒅𝒙 Let I1=∫_0^(𝜋/2 )▒𝑙𝑜𝑔(𝑠𝑖𝑛𝑥) 𝑑𝑥 Solving 𝐈𝟏 I1=∫_0^(𝜋/2 )▒𝑙𝑜𝑔(𝑠𝑖𝑛𝑥) 𝑑𝑥 ∴ I1=∫_𝟎^(𝝅/𝟐)▒𝒔𝒊𝒏(𝝅/𝟐−𝒙)𝒅𝒙 I1= ∫_𝟎^(𝝅/𝟐)▒𝒍𝒐𝒈(𝒄𝒐𝒔⁡𝒙 )𝒅𝒙 Adding (2) and (3) i.e. (2) + (3) 𝐈𝟏 + 𝐈𝟏 =∫_𝟎^(𝝅/𝟐)▒〖𝒍𝒐𝒈(𝒔𝒊𝒏⁡𝒙 )𝒅𝒙+∫_𝟎^(𝝅/𝟐)▒𝒍𝒐𝒈(𝐜𝐨𝐬⁡𝒙 )𝒅𝒙〗 "Using" 𝒍𝒐𝒈⁡𝒂 + 𝒍𝒐𝒈⁡𝒃 = 𝒍𝒐𝒈⁡(𝒂.𝒃) 2I1 =∫_𝟎^(𝝅/𝟐)▒〖𝐥𝐨𝐠⁡[𝐬𝐢𝐧⁡〖𝒙 𝐜𝐨𝐬⁡𝒙 〗 ] 𝒅𝒙〗 2I1 = ∫_𝟎^(𝝅/𝟐)▒〖𝒍𝒐𝒈⁡[𝟐𝒔𝒊𝒏⁡〖𝒙 𝒄𝒐𝒔⁡𝒙 〗/𝟐] 𝒅𝒙〗 "Using " 𝒍𝒐𝒈(𝒂/𝒃) = log⁡(𝑎) – log⁡(𝑏)) 2I1 = ∫_0^(𝜋/2)▒[log[2sin⁡〖𝑥 cos⁡𝑥 〗 ]−𝑙𝑜𝑔(2)]𝑑𝑥 "Using" 𝒔𝒊𝒏⁡𝟐𝒙=2 sin⁡〖𝑥 cos⁡𝑥 〗 2I1 = ∫_0^(𝜋/2)▒[log[sin⁡2𝑥 ]−𝑙𝑜𝑔2]𝑑𝑥 2I1 = ∫_𝟎^(𝝅/𝟐)▒𝐥𝐨𝐠[𝒔𝒊𝒏⁡𝟐𝒙 ]𝒅𝒙−∫_0^(𝜋/2)▒log(2)𝑑𝑥 Solving 𝐈𝟐 𝐈𝟐=∫_𝟎^(𝝅/𝟐)▒〖𝐥𝐨𝐠 𝒔𝒊𝒏⁡𝟐𝒙 𝒅𝒙〗 Let 2𝑥=𝑡 Differentiating both sides w.r.t.𝑥 2=𝑑𝑡/𝑑𝑥 𝑑𝑥=𝑑𝑡/2 ∴ Putting the values of t and 𝑑𝑡 and changing the limits, I2 =∫_0^(𝜋/2)▒log(sin⁡2𝑥 )𝑑𝑥 I2 = ∫_0^𝜋▒〖log(sin⁡𝑡 ) 𝑑𝑡/2〗 𝐈𝟐 = 𝟏/𝟐 ∫_𝟎^𝝅▒𝐥𝐨𝐠(𝒔𝒊𝒏⁡𝒕 )𝒅𝒕 Here, 𝑓(𝑡)=log⁡𝑠𝑖𝑛𝑡 𝑓(2𝑎−𝑡)=𝑓(2𝜋−𝑡)=log⁡𝑠𝑖𝑛(2𝜋−𝑡)=log⁡sin⁡𝑡 As, 𝒇(𝒕)=𝒇(𝟐𝒂−𝒕) ∴ 𝐈𝟐 = 1/2 ∫_0^𝜋▒log⁡sin⁡〖𝑡 𝑑𝑡〗 =𝟏/𝟐 × 𝟐∫_𝟎^(𝝅/𝟐)▒𝒍𝒐𝒈⁡𝒔𝒊𝒏⁡〖𝒕. 𝒅𝒕〗 =∫_0^(𝜋/2)▒log⁡sin⁡〖𝑡. 𝑑𝑡〗 =∫_0^(𝜋/2)▒log⁡sin⁡〖𝑥 𝑑𝑥〗 Putting the value of I2 in equation (3), we get 2I1 =∫_𝟎^(𝝅/𝟐)▒𝐥𝐨𝐠[𝒔𝒊𝒏⁡𝟐𝒙 ]⁡𝒅𝒙 −∫_0^(𝜋/2)▒log(2)⁡𝑑𝑥 2I1 = ∫_𝟎^(𝝅/𝟐)▒𝐥𝐨𝐠(𝒔𝒊𝒏⁡𝒙 )⁡𝒅𝒙 −log(2) ∫_𝟎^(𝝅/𝟐)▒〖𝟏.〗⁡𝒅𝒙 2I1 = I1 − log(2) [𝑥]_0^(𝜋/2) 2I1−I1=−log⁡2 [𝜋/2−0] 𝐈𝟏=−𝒍𝒐𝒈⁡𝟐 [𝝅/𝟐] Hence, 𝐈=2 I1= 2 × (−𝜋)/2 log⁡2 =−𝝅 𝒍𝒐𝒈⁡𝟐