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

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

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Answer 1 · 2016-08-31T10:59:03

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

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.!