Given, Let dx =2t dt Now, Consider, t =sin k dt=cos k dk =2 ∫cos 2 k dk = ∫2 cos 2 k dk = ∫cos 2k-1 dk [since, cos 2x=2cos 2 x-1] =t cos(sin -1 t) -2sin -1 t+2c =√x cos(sin -1 √x )-2 sin -1 √x+2c

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