OK my friends I am giving the solution First break cos2x as 2cos²x - 1

We get

_{[ ]}^{[ ]} (23cos²x-3-1)/cosx  dx

= _{[ ]}^{[ ]} 23cosx dx - (3+1)_{[ ]}^{[ ]} secx dx

=23sinx |₀^/6 - (1+ 3)ln(secx + tanx) |₀^/6

⁼3 - (1+3)ln3

I hope now u r satisfied........

I hopw I can expect  some points now......