Ex 7.11, 16

Transcript

Ex 7.11, 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﷯ 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﷯ 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﷯ ∴ 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﷯﷯ 2I1 = 0﷮ 𝜋﷮2﷯﷮ log﷮ sin﷮𝑥 cos﷮𝑥﷯﷯﷯﷯𝑑𝑥﷯ 2I1 = 0﷮ 𝜋﷮2﷯﷮ log﷮ 2sin﷮𝑥 cos﷮𝑥﷯﷯﷮2﷯﷯﷯𝑑𝑥﷯ 2I1 = 0﷮ 𝜋﷮2﷯﷮ log 2sin﷮𝑥 cos﷮𝑥﷯﷯﷯−𝑙𝑜𝑔 2﷯﷯𝑑𝑥﷯ 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﷯﷯﷯ I2= 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﷯﷯ ∴ 𝐼1= − 𝜋﷮2﷯ log﷮2﷯ Hence, 𝐈=2 I1= 2 × −𝜋﷮2﷯ log﷮2﷯=−𝝅 𝒍𝒐𝒈﷮𝟐﷯