Answer :

Given,

Let




dx =2t dt


Now,



Consider, t=sin k


dt=cos k dk




=2∫cos2k dk


=∫2 cos2k dk


=∫cos 2k-1 dk [since, cos 2x=2cos2x-1]




=t cos(sin-1 t) -2sin-1 t+2c


=√x cos(sin-1√x )-2 sin-1√x+2c


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