Question

Find the integral of arcsec(x)/(x^6). I arrived at the integral 1/(sec^5) with integration by parts and trig substitution but now I don't know what to do.

Answer 1

That's rough, I really don't how to to figure that one out. Sorry.

Answer 2

now, 1/(secx^5)=(1/secx)^5=cosx^5
so the integral of 1/(sec^5) is the same as the integral of cos^5
=int (cosx^4*cosx) dx
=int [[(1-sinx^2)^2]*cosx] dx
use substitution again, let u=sinx => du=cosdx
then we have
=int [(1-u^2)^2)du=int(1-2u^2+u^4)du
=u-2u^3/3+u^5/5
=sinx-2sinx^3/3+sinx^5/5

Answer 3

thank you very much!