Hint:

1 + ln(cosy)^(3/2) = 1+(3/2)ln(cosy)let u = 1+(3/2)ln(cosy)

Consider T = ln{ln[(cosy)^(3/2) + 1]}

dT/dy = (-3/2)* tan y/{ln[(cosy)^(3/2) + 1]}

so, I = ∫tan y/{ln[(cosy)^(3/2) + 1]}dy = -2/3T + C

I = (-2/3)ln{ln[(cosy)^(3/2) + 1]} + C