Verified answer

(tan(x))^4 = tan²(x)(sec²(x) - 1)

= sec²(x)tan²(x) - tan²(x)

= sec²(x)tan²(x) - (sec²(x) - 1)

= sec²(x)tan²(x) - sec²(x) + 1

∫tan^4 xdx = ∫sec²(x)tan²(x)dx - ∫sec²(x) dx + ∫ 1 dx

= ∫sec²(x)tan²(x) dx - tan(x) + x + C

to do sec²(x)tan²(x)

sub u = tan(x)

du = sec²(x) dx

∫sec²(x)tan²(x) dx = ∫ u² du

= u³/3 + C

= tan³(x)/3 + C

In all

∫tan^4 xdx = tan³(x)/3 - tan(x) + x + C

Although this is NOT the answer that wolfram gives, so I am wondering where I have made a mistake. But I believe my method is good.

Edit:

But I now see that someone else gave the same answer, so maybe it is good after all!

Break it up as tan^2(x) tan^2(x)

s^2+c^2 =1

tan^2(x) = sec^2(x) -1

Answer = int tan^2(x)sec^2(x) - tan^2(x) dx

Now split up the integral as a sum of intergals

= int [tan^2(x)sec^2(x)]dx - int [ tan^2(x)] dx

For the first integral sub u=tan(x) and for the second replace tan^2(x) with sec^2(x) -1 and intergate term by term.

Final answer= (1/3) tan^3 (x) - [ tan(x) -x] +C = (1/3) tan^3(x)-tan(x)+x +C