Evaluate `int tan^(3)x dx`

Answer

`(1)/(2) tan^(2)x-log|sec x|+C`

Explanation

`I=int tan^(3)x dx`
`=int tan^(2)x tanx dx`
`=int (sec^(2)x-1)tanx dx`
`=int tanx sec^(2) x dx - int tanx dx`
`=I_(1)-log|secx|+C, " where " I_(1)=int tanx sec^(2)x dx`
Putting `tan x =t " and " sec^(2) x dx=dt " in " I_(1),` we get
`I_(1)= int t dt = (t^(2))/(2)=(1)/(2) tan^(2)x +C`
Hence, `I=(1)/(2) tan^(2)x-log|sec x|+C`.