# Ex 7.6, 22

Last updated at Dec. 8, 2016 by Teachoo

Last updated at Dec. 8, 2016 by Teachoo

Transcript

Ex 7.6, 22 sin 1 2 1 + 2 Simplifying the given function sin 1 2 1 + 2 Put =tan = tan 1 sin 1 2 1 + 2 = sin 1 2 tan 1 + tan 2 = sin 1 sin 2 = 2t =2 tan 1 Thus, our function becomes sin 1 2 1 + 2 = 2 tan 1 =2 tan 1 1. = 2 tan 1 1 . 2 tan 1 1 . = 2 tan 1 2 1 1 + 2 . . = 2 tan 1 2 1 + 2 . Solving I1 I1 = 1 + 2 . Let 1 + 2 = Differentiating both sides . . . 0 + 2 = = 2 Now, I1 = 1 + 2 . Putting the value of 1+ 2 = t and = 2 , we get I1 = . 2 I1 = 1 2 1 . I1 = 1 2 log + 1 I1 = 1 2 log 1+ 2 + 1 Putting the value of I1 in (1) , tan 1 . =2 tan 1 2 1 + 2 . =2 tan 1 2 1 2 log 1+ 2 + 1 =2 tan 1 log 1+ 2 2 1 = + +

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About the Author

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.