# Ex 7.11, 16

Last updated at Dec. 8, 2016 by Teachoo

Last updated at Dec. 8, 2016 by Teachoo

Transcript

Ex 7.11, 16 By using the properties of definite integrals, evaluate the integrals : 0𝜋 log 1+ cos𝑥 𝑑𝑥 Let I= 0𝜋 log 1+ cos𝑥 𝑑𝑥 ∴ I= 0𝜋𝑙𝑜𝑔 1+𝑐𝑜𝑠 𝜋−𝑥 𝑑𝑥 I= 0𝜋𝑙𝑜𝑔 1− cos𝑥 𝑑𝑥 Adding (1) and (2) i.e. (1) + (2) I+I= 0𝜋𝑙𝑜𝑔 1+ cos𝑥𝑑𝑥+ 0𝜋𝑙𝑜𝑔 1− cos𝑥𝑑𝑥 2I= 0𝜋 𝑙𝑜𝑔 1+ cos𝑥+𝑙𝑜𝑔 1− cos𝑥𝑑𝑥 2I= 0𝜋𝑙𝑜𝑔 1+ cos𝑥 1− cos𝑥𝑑𝑥 2I= 0𝜋𝑙𝑜𝑔 1− cos2𝑥𝑑𝑥 2I= 0𝜋𝑙𝑜𝑔 𝑠𝑖𝑛2𝑥𝑑𝑥 2I= 0𝜋2 𝑙𝑜𝑔 𝑠𝑖𝑛𝑥𝑑𝑥 2I=2 0𝜋𝑙𝑜𝑔 𝑠𝑖𝑛𝑥𝑑𝑥 I= 0𝜋𝑙𝑜𝑔 𝑠𝑖𝑛𝑥𝑑𝑥 Here, 𝑓 𝑥= log sin𝑥 f 𝜋−𝑥=𝑙𝑜𝑔 𝑠𝑖𝑛 𝜋−𝑥𝑑𝑥 =𝑙𝑜𝑔 sin𝑥𝑑𝑥 =𝑓 𝑥 ∴ I= 0𝜋𝑙𝑜𝑔 sin𝑥𝑑𝑥=2 0 𝜋2𝑙𝑜𝑔 sin𝑥𝑑𝑥 Let I1= 0 𝜋2 𝑙𝑜𝑔 𝑠𝑖𝑛𝑥𝑑𝑥 Solving 𝐈𝟏 I1= 0 𝜋2 𝑙𝑜𝑔 𝑠𝑖𝑛𝑥𝑑𝑥 ∴ I1= 0 𝜋2𝑠𝑖𝑛 𝜋2−𝑥𝑑𝑥 I1= 0 𝜋2𝑙𝑜𝑔 cos𝑥𝑑𝑥 Adding (2) and (3) i.e. (2) + (3) I1 + I1 = 0 𝜋2𝑙𝑜𝑔 sin𝑥𝑑𝑥+ 0 𝜋2𝑙𝑜𝑔 cos𝑥𝑑𝑥 2I1 = 0 𝜋2 log sin𝑥 cos𝑥𝑑𝑥 2I1 = 0 𝜋2 log 2sin𝑥 cos𝑥2𝑑𝑥 2I1 = 0 𝜋2 log 2sin𝑥 cos𝑥−𝑙𝑜𝑔 2𝑑𝑥 2I1 = 0 𝜋2 log sin2𝑥−𝑙𝑜𝑔2𝑑𝑥 2I1 = 0 𝜋2log sin2𝑥𝑑𝑥− 0 𝜋2log 2𝑑𝑥 Solving 𝐈𝟐 I2= 0 𝜋2log sin2𝑥𝑑𝑥 Let 2𝑥=𝑡 Differentiating both sides w.r.t.𝑥 2= 𝑑𝑡𝑑𝑥 𝑑𝑥= 𝑑𝑡2 ∴ Putting the values of t and 𝑑𝑡 and changing the limits, I2 = 0 𝜋2log sin2𝑥𝑑𝑥 I2 = 0𝜋log sin𝑡 𝑑𝑡2 I2 = 12 0𝜋log sin𝑡𝑑𝑡 Here, 𝑓 𝑡= log𝑠𝑖𝑛𝑡 𝑓 2𝑎−𝑡=𝑓 2𝜋−𝑡= log𝑠𝑖𝑛 2𝜋−𝑡= log sin𝑡 As, 𝑓 𝑡=𝑓 2𝑎−𝑡 ∴ I2 = 12 0𝜋 log sin𝑡 𝑑𝑡= 12 ×2 0 𝜋2 log sin𝑡. 𝑑𝑡 = 0 𝜋2 log sin𝑡. 𝑑𝑡 I2= 0 𝜋2 log sin𝑥 𝑑𝑥 Putting the value of I2 in equation (3), we get 2I1 = 0 𝜋2 log sin2𝑥𝑑𝑥− 0 𝜋2 log 2𝑑𝑥 2I1 = 0 𝜋2 log sin𝑥𝑑𝑥−log 2 0 𝜋2 1.𝑑𝑥 2I1 = I1 − log 2 𝑥0 𝜋2 2I1−I1=−log2 𝜋2−0 I1=−log2 𝜋2 ∴ 𝐼1= − 𝜋2 log2 Hence, 𝐈=2 I1= 2 × −𝜋2 log2=−𝝅 𝒍𝒐𝒈𝟐

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