Integrate sin^5 x cos^-2 x dx

The answer is sec x + 2cos x - cos^3 x/3 + c

I need to see how this answer is attained because I can't get it.

∫ sin^5(x) dx(cos x)^(-2)

=> ∫ sin^5(x) dx / cos^2(x)

= ∫ sin^4(x) sin x dx / cos^2(x)

= ∫ ( 1 - cos^2(x))^2 sin x dx / cos^2(x)

let cos x = t
- sin x dx = dt
sin x dx = - dt

now the integral changes to


- ∫ ( 1 - t^2)^2 dt /t^2

= - ∫ ( 1 + t^4 - 2t^2 ) dt /t^2

= -∫ dt /t^2 - ∫ t^4 dt /t^2 + 2∫ t^2 /t^2 dt

= -∫ dt /t^2 - ∫ t^2 dt + 2∫ dt

= (1 / t) - (1/3)t^3 + 2t + C

back substitute t = cos x

(1 / cos x) - (1/3)cos^3(x) + 2 cos x + C

= sec x - (1/3)cos^3(x) + 2 cos x + C