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
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 You are here
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 April 16, 2024 by Teachoo
Ex 7.10, 16 By using the properties of definite integrals, evaluate the integrals : ∫_0^𝜋▒log(1+cos𝑥 ) 𝑑𝑥 Let I=∫_𝟎^𝝅▒𝒍𝒐𝒈(𝟏+𝒄𝒐𝒔𝒙 ) 𝒅𝒙 ∴ I=∫_0^𝜋▒log(1+𝑐𝑜𝑠(𝜋−𝑥)) 𝑑𝑥 𝐈=∫_𝟎^𝝅▒𝐥𝐨𝐠(𝟏−𝒄𝒐𝒔𝒙 ) 𝑑𝑥 Adding (1) and (2) i.e. (1) + (2) I+I=∫_0^𝜋▒𝑙𝑜𝑔(1+cos𝑥 )𝑑𝑥+∫_0^𝜋▒𝑙𝑜𝑔(1−cos𝑥 )𝑑𝑥 𝟐𝐈=∫_𝟎^𝝅▒[𝒍𝒐𝒈(𝟏+𝒄𝒐𝒔𝒙 )+𝒍𝒐𝒈(𝟏−𝒄𝒐𝒔𝒙 )]𝒅𝒙 "Using log(a) + log(b)" = "log(a.b)") 2I=∫_0^𝜋▒𝒍𝒐𝒈[(𝟏+𝒄𝒐𝒔𝒙 )(𝟏−𝒄𝒐𝒔𝒙 )]𝒅𝒙 2I=∫_0^𝜋▒𝑙𝑜𝑔[1−cos^2𝑥 ]𝑑𝑥 2I=∫_0^𝜋▒𝑙𝑜𝑔(〖𝑠𝑖𝑛〗^2𝑥 )𝑑𝑥 Using log 𝒂^𝒃=𝒃 𝒍𝒐𝒈𝒂 2I=∫_0^𝜋▒〖2 𝑙𝑜𝑔(𝑠𝑖𝑛𝑥 )𝑑𝑥〗 2I=2∫_0^𝜋▒𝑙𝑜𝑔(𝑠𝑖𝑛𝑥 )𝑑𝑥 𝐈=∫_𝟎^𝝅▒𝒍𝒐𝒈(𝒔𝒊𝒏𝒙 )𝒅𝒙 Here, 𝒇(𝒙)=logsin𝑥 f(𝝅−𝒙)=𝑙𝑜𝑔[𝑠𝑖𝑛(𝜋−𝑥)]𝑑𝑥 =𝑙𝑜𝑔(sin𝑥 )𝑑𝑥 =𝒇(𝒙) Therefore, I=∫_𝟎^𝝅▒𝒍𝒐𝒈(𝐬𝐢𝐧𝒙 )𝒅𝒙=𝟐∫_𝟎^(𝝅/𝟐)▒𝒍𝒐𝒈(𝒔𝒊𝒏𝒙 )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(𝒄𝒐𝒔𝒙 )𝒅𝒙 Adding (2) and (3) i.e. (2) + (3) 𝐈𝟏 + 𝐈𝟏 =∫_𝟎^(𝝅/𝟐)▒〖𝒍𝒐𝒈(𝒔𝒊𝒏𝒙 )𝒅𝒙+∫_𝟎^(𝝅/𝟐)▒𝒍𝒐𝒈(𝐜𝐨𝐬𝒙 )𝒅𝒙〗 "Using" 𝒍𝒐𝒈𝒂 + 𝒍𝒐𝒈𝒃 = 𝒍𝒐𝒈(𝒂.𝒃) 2I1 =∫_𝟎^(𝝅/𝟐)▒〖𝐥𝐨𝐠[𝐬𝐢𝐧〖𝒙 𝐜𝐨𝐬𝒙 〗 ] 𝒅𝒙〗 2I1 = ∫_𝟎^(𝝅/𝟐)▒〖𝒍𝒐𝒈[𝟐𝒔𝒊𝒏〖𝒙 𝒄𝒐𝒔𝒙 〗/𝟐] 𝒅𝒙〗 "Using " 𝒍𝒐𝒈(𝒂/𝒃) = log(𝑎) – log(𝑏)) 2I1 = ∫_0^(𝜋/2)▒[log[2sin〖𝑥 cos𝑥 〗 ]−𝑙𝑜𝑔(2)]𝑑𝑥 "Using" 𝒔𝒊𝒏𝟐𝒙=2 sin〖𝑥 cos𝑥 〗 2I1 = ∫_0^(𝜋/2)▒[log[sin2𝑥 ]−𝑙𝑜𝑔2]𝑑𝑥 2I1 = ∫_𝟎^(𝝅/𝟐)▒𝐥𝐨𝐠[𝒔𝒊𝒏𝟐𝒙 ]𝒅𝒙−∫_0^(𝜋/2)▒log(2)𝑑𝑥 Solving 𝐈𝟐 𝐈𝟐=∫_𝟎^(𝝅/𝟐)▒〖𝐥𝐨𝐠 𝒔𝒊𝒏𝟐𝒙 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(sin2𝑥 )𝑑𝑥 I2 = ∫_0^𝜋▒〖log(sin𝑡 ) 𝑑𝑡/2〗 𝐈𝟐 = 𝟏/𝟐 ∫_𝟎^𝝅▒𝐥𝐨𝐠(𝒔𝒊𝒏𝒕 )𝒅𝒕 Here, 𝑓(𝑡)=log𝑠𝑖𝑛𝑡 𝑓(2𝑎−𝑡)=𝑓(2𝜋−𝑡)=log𝑠𝑖𝑛(2𝜋−𝑡)=logsin𝑡 As, 𝒇(𝒕)=𝒇(𝟐𝒂−𝒕) ∴ 𝐈𝟐 = 1/2 ∫_0^𝜋▒logsin〖𝑡 𝑑𝑡〗 =𝟏/𝟐 × 𝟐∫_𝟎^(𝝅/𝟐)▒𝒍𝒐𝒈𝒔𝒊𝒏〖𝒕. 𝒅𝒕〗 =∫_0^(𝜋/2)▒logsin〖𝑡. 𝑑𝑡〗 =∫_0^(𝜋/2)▒logsin〖𝑥 𝑑𝑥〗 Putting the value of I2 in equation (3), we get 2I1 =∫_𝟎^(𝝅/𝟐)▒𝐥𝐨𝐠[𝒔𝒊𝒏𝟐𝒙 ]𝒅𝒙 −∫_0^(𝜋/2)▒log(2)𝑑𝑥 2I1 = ∫_𝟎^(𝝅/𝟐)▒𝐥𝐨𝐠(𝒔𝒊𝒏𝒙 )𝒅𝒙 −log(2) ∫_𝟎^(𝝅/𝟐)▒〖𝟏.〗𝒅𝒙 2I1 = I1 − log(2) [𝑥]_0^(𝜋/2) 2I1−I1=−log2 [𝜋/2−0] 𝐈𝟏=−𝒍𝒐𝒈𝟐 [𝝅/𝟐] Hence, 𝐈=2 I1= 2 × (−𝜋)/2 log2 =−𝝅 𝒍𝒐𝒈𝟐