Answer :

Let


Now, we know that, sin2x + cos2x = 1


Therefore,



Also, sin2x = 2.sinx.cosx


Therefore,





Now, we know that,


Hence,


OR


Let


Let sin-12x = t, 2x = sint


Differentiating both sides, we get,


2dx = cost dt


Therefore,



Now, Integrating, by parts, taking t as 1st function and cost as the second function.


We know that,



Therefore,




Putting the value of t, we get,



Hence,


I = x.sin-12x – x + C


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