Verified answer

∫ sin(x)*(sec(x))^9 dx

∫ sin(x)*1/(cos(x))^9 dx

let u = cos(x)

du = -sin(x) dx

-du = sin(x) dx

∫ -1/u^9 du

-∫ u^-9 du

-(-1/8) u^-8

1/(8*u^8)

1/(8*(cos(x))^9))

or

1/8 * (sec(x))^8