Question

How do you evaluate the integral `int sec^3x/tanx`?

Answer 1

`1/2ln|(cosx-1)/(cosx+1)|+secx+C, or, ln|tan(x/2)|+secx+C`.

Explanation:

Let `I=intsec^3x/tanxdx=int(1/cos^3x)(cosx/sinx)dx`

`=int1/(cos^2xsinx)dx=intsinx/(cos^2xsin^2x)dx`

`:. I=-int{(-sinx)/{cos^2x(1-cos^2x)}dx`

Substituting `cosx=t," so that, "-sinxdx=dt`, we get,

`I=int1/{t^2(t^2-1)}dt=int{t^2-(t^2-1)}/{t^2(t^2-1)}dt`

`=int[t^2/{t^2(t^2-1)}-(t^2-1)/{t^2(t^2-1)}]dt`

`=int[1/(t^2-1)-1/t^2]dt`

`1/2ln|(t-1)/(t+1)|+1/t`.

Since, `t=cosx`, we have,

`I=1/2ln|(cosx-1)/(cosx+1)|+secx+C`.

Enjoy Maths.!

N.B.:-`I` can further be simplified as `ln|tan(x/2)|+secx+C`.