Question

How do you evaluate the integral of `int 1/ (sec^3 x tan x) dx`?

Answer 1

`=int1/(sec^3x(sinx/cosx))dx=intsinx/(sec^4x xxsin^2x)dx=int(cos^4xxsinx)/(1-cos^2x)dx`

proceed putting
`cosx = z and dz= -sinxdx`
`I=-intz^4/(1-z^2)dz=intz^4/(z^2-1)dz`

try now

Answer 2

`int1/(sec^3x tanx) dx =ln|cscx-cotx|+cosx+1/3 cos^3x +C`

Explanation:

`int1/(sec^3x tanx) dx = int1/sec^3x * 1/tanx dx=int cos^3xcot x*dx`

`=intcos^3x *cosx/sinx *dx = intcos^4x/sinx dx=int(cos^2x*cos^2x)/sinx dx`

`=int( (1-sin^2x)(1-sin^2x))/sinx dx =int(1-2sin^2x+sin^4x)/sinx dx`

`=int (1/sinx -2sin^2x/sinx +sin^4x /sinx) dx`

`=int (cscx -2sinx +sin^3x)dx`
`= int(cscx-2sinx+sinxsin^2x)dx`
`= int(cscx-2sinx+sinx(1-cos^2x))dx`
`= int(cscx-2sinx+sinx-cos^2x sin x)dx`
`= int(cscx-sinx-cos^2x sin x)dx`
`=ln|cscx-cotx|+cosx+1/3 cos^3x +C`