Example 36 - Chapter 7 Class 12 Integrals
Last updated at July 19, 2019 by Teachoo
Transcript
Example 36 Evaluate 0 2 log sin Let I1= 0 2 I1= 0 2 2 I1= 0 2 cos Adding (1) and (2) i.e. (1) + (2) I1 + I1 = 0 2 sin + 0 2 cos 2I1 = 0 2 log sin cos 2I1 = 0 2 log 2sin cos 2 2I1 = 0 2 log 2sin cos 2 2I1 = 0 2 log sin 2 2 2I1 = 0 2 log sin 2 0 2 log 2 Solving I2= 0 2 log sin 2 Let 2 = Differentiating both sides w.r.t. 2= = 2 Putting the values of t and and changing the limits, I2 = 0 2 log sin 2 I2 = 0 log sin 2 I2 = 1 2 0 log sin Here, = log 2 = 2 = log 2 = log sin As, = 2 I2 = 1 2 0 log sin = 1 2 2 0 2 log sin . = 0 2 log sin . I2= 0 2 log sin Putting the value of I2 in equation (3), we get 2I1 = 0 2 log sin 2 0 2 log 2 2I1 = 0 2 log sin log 2 0 2 1. 2I1 = I1 log 2 0 2 2I1 I1= log 2 2 0 I1= log 2 2 =
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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.