What is #int 1/(t^3sqrt( t^2 - 1))dt#?

Question

5329 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-07-04T19:48:00

#\int \frac{1}{t^3\sqrt{t^2-1}}dt=1/2\sec^{-1}(t)+1/4\sin(2\sec^{-1}(t))+C#

Let #t=\sec\theta\implies dt=\sec\theta\tan\theta\ d\theta#

#\int \frac{1}{t^3\sqrt{t^2-1}}dt#

#=\int \frac{1}{\sec^3\theta\sqrt{sec^2\theta-1}}(\sec\theta\tan\theta\ d\theta)#

#=\int \frac{\tan\theta}{\sec^2\theta\tan\theta} d\theta#

#=\int \frac{1}{\sec^2\theta} d\theta#

#=\int \cos^2\theta\ d\theta#

#=\int \frac{1+\cos2\theta}{2}\ d\theta#

#=1/2\int d\theta+1/2\int \cos2\theta\ d\theta#

#=1/2\theta+1/2(1/2)\sin2\theta+C#

#=1/2\sec^{-1}(t)+1/4\sin(2\sec^{-1}(t))+C#