Recall that

`sec^2x=1+tan^2x`

So we can write the part under the integral as:

`sqrt(tan\ x)\ sec^4 x =(tan^(1//2) x)\ sec^2 x\ sec^2 x`

`=(tan^(1//2) x )(1+tan^2 x )sec^2 x`

`=(tan^(1//2) x+tan^(5//2)x)sec^2 x`

Next, we put `u = tan\ x`, giving `du = sec^2x\ dx`

So our integral becomes:

`int (tan^(1//2) x+tan^(5//2)x)sec^2 x\ dx =int(u^(1//2)+u^(5//2))du`

`=2/3u^(3//2)+2/7u^(7//2)+K`

`=2/3tan^(3//2)x+2/7tan^(7//2)x+K`