Use Integration by Parts,

Let,

u=3x and v'=sin x/cos^4 x

Then,

u'=3 and v=INTEGRAL sin x/cos ^4 x dx

But how do we find that integral?

We have, (multiply by negative)

-INTEGRAL -sin x/cos^4 x dx

Let, t=cos x then t'=-sin x

Apply substitution rule,

-INTEGRAL t'/t^4 dx=INTEGRAL 1/t^4 dt

Thus, by using t^{-4} and exponent rule we have,

-1/3*t^{-3}

Substitute back,

(-1/3)cos^{-3} x

Thus by parts,

-xcos^3 t+(1/3)INTEGRAL 1/cos^3 x dx

The problem is how to deal with that integral.

I believe that if you can express it then you should write,

sec x*sec^x

Apply integration by parts.