Evaluate
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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