Evaluate: `intsec^3x\ dx`

Explanation

Let `I=int sec^(3)x dx " d"=int secx sec^(2)x dx`
`=int sqrt(1+tan^(2)x)sec^(2)x dx`
Put `tan x=z " and " sec^(2)x dx=dz`
` :. I=int sqrt(1+z^(2))dz`
`=(zsqrt (z^(2)+1))/(2)+(1)/(2)log|z+sqrt(z^(2)+1)|+C`
` =(tanx sec x)/(2)+(1)/(2)log|tanx +secx| +C`
`=(1)/(2)[secx tanx + log|secx +tanx|] +C`