How do you integrate int 1/(t^3sqrt(t^2-1)) by trigonometric substitution?

How do you integrate int 1/(t^3sqrt(t^2-1))dt by trigonometric substitution?

1 Answer
Aug 31, 2016

1/2(arc sect+sqrt(t^2-1)/t^2)+C.

Explanation:

Let I=int1/(t^3sqrt(t^2-1))dt.

We subst. t=sec x, "so that," dt=sec xtan xdx. Hence,

I=int(secxtanx)/(sec^3xtanx)dx=intcos^2xdx=int(1+cos2x)/2dx

=1/2{x+(sin2x)/2}=1/2(x+sinxcos)

Now, secx=t rArr x=arc sect, cos x=1/t, sinx=sqrt(1-1/t^2).

Therefore, I=1/2{arc sect+(1/t)(sqrt(t^2-1)/t}, or,

I=1/2(arc sect+sqrt(t^2-1)/t^2)+C.

Enjoy Maths.!