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

1 Answer

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

Explanation:

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