How can I evaluate these without integration tables?

Ok, Let's go with the first one :

Int[(tanx)^3(secx)^3]dx

Remember : tan^2x + 1 = sec^2x

int[tanx*(sec^2x - 1)*sec*3x]dx

int[tanx*(sec^5x - sec^3x)]dx

remember : tanx = sinx / cosx

secx = 1 / cosx

int[(sinx / cos^6x)dx - int[sinx / cox^4x]dx

Now we have easier integrals, remember : - d(cosx) = sinx

-int[dcosx / cos^6x] + int[dcosx / cos^4x]dx

1/5*(cosx)^-5 - 1/3*(cosx)^-3

That's it

Second problem :

Int[(cos(3x))^4]dx

Remember :

2*cos^2(3x) = cos6x + 1

cos^2(3x) = (cos6x + 1) / 2

cos^4(3x) = ( 1 + cos^2(6x) + 2*cos6x ) / 4

Let's replace it on the integral :

int[ ( 1 + cos^2(6x) + 2*cos6x ) / 4)]dx

1/4*int[dx] + 1/4*int[cos^2(6x)] + 1/2*int(cos6x)

1/4*x + 1/12*sin(6x) + 1/4*int[cos^2(6x)]dx.......(**)

This : 1/4*int[cos^2(6x)]dx is simple :

1/8*int[2cos^2(6x)]dx

1/8*int[cos(12x) + 1]dx

1/64*sin(12x) + x/8 ..... (*)

Let's replace : (*) on (**)

1/4*x + 1/12*sin(6x) + 1/64*sin(12x) + x/8

That's it

Hope that helps