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Ex 7.11, 16 - Evaluate definite integral log (1 + cos x) dx - Definate Integration by properties - P4

Ex 7.11, 16 - Chapter 7 Class 12 Integrals - Part 2
Ex 7.11, 16 - Chapter 7 Class 12 Integrals - Part 3 Ex 7.11, 16 - Chapter 7 Class 12 Integrals - Part 4 Ex 7.11, 16 - Chapter 7 Class 12 Integrals - Part 5 Ex 7.11, 16 - Chapter 7 Class 12 Integrals - Part 6 Ex 7.11, 16 - Chapter 7 Class 12 Integrals - Part 7 Ex 7.11, 16 - Chapter 7 Class 12 Integrals - Part 8

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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﷯

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.