Integral of sec squared x times tan squared x

this is the only problem i cant do in the
past papers for the test tomorrow. please help:)

∫ sec²(x)·tan²(x) dx = ?

Use u-substitution:
let u = tan(x) and du = sec²(x) dx

∫ sec²(x)·tan²(x) dx = ∫ u² du
∫ sec²(x)·tan²(x) dx = u³/3+c
∫ sec²(x)·tan²(x) dx = tan³(x)/3+c

Solution: F(x) = tan³(x)/3+c

∫ sec^2(x) tan^2(x) dx

Let u = tan(x), then du = sec^2(x) dx, and the integral becomes:

∫ u^2 du = (u^3) / 3 + C = [tan^3(x) / 3] + C