Evaluate the indefinite integral.

Integral (tan^8xsec^2x) dx

u = tan(x)
du = sec(x)^2 * dx

tan(x)^8 * sec(x)^2 * dx =>
u^8 * du

Integrate

(1/9) * u^9 + C

Back substitute

(1/9) * tan(x)^9 + C

∫ (tan^8xsec^2x) dx =
∫ tan^8x d(tanx) = (1/9)tan^9 (x) + C

Or use
tan(x) = u
sec^2 x dx = du
∫ (tan^8xsec^2x) dx becomes
∫ u^8 du = (1/9) u^9 + C
back to x
(1/9)tan^9 (x) + C