Int (sinx)^4/(cosx)6

this is  (sinx/cosx)^4 / cosx ^2

= (tanx)^4 * (secx)^2

put tanx = t

diff. sec^2x dx = dt

we get  (t)^4 dt    ( sec^2x dx = dt  and tanx = t)

 = t^5/5

= tanx^5 / 5

int tan x⁴ .sec x²  dx

=int tan x⁴ (1+tan x²) dx

=int tan x⁴ dx +int tan x⁶ dx                  it is the formula

=(tan x³⁾ /3 -tan x +(tan x⁵ )/5 -(tan x^{3  )}/3 +tan x

=(tan x⁵) /5