Question

Evaluate the Integral\(\int {\frac{{dt}}{{{t^2}\sqrt {{t^2} - 16} }}} \)

Short answer

Given an indefinite integration

Underneath the roots, there is the form \(\sqrt {{t^2} + {a^2}} \)where a>0 such can be integrated by taking \(t = a\sec \theta \)

Step-by-step solution

Step-1:- Given Data

Consider \(t = a\sec \theta \) i.e., \(t = 4\sec \theta \)

\(dt = a\sec \theta \tan \theta d\theta \) i.e., a=4

Step-2:- Substitute t and dt in the expression 

\(\int {\frac{{dt}}{{{t^2}\sqrt {{t^2} - 16} }}} \)\(\)=\(\int {\frac{{4\sec \theta \tan \theta d\theta }}{{{{\left( {4\sec \theta } \right)}^2}\sqrt {{{\left( {4\sec \theta } \right)}^2} - {{\left( 4 \right)}^2}} }}} \)

= \(\int {\frac{{4\sec \theta \tan \theta d\theta }}{{{{(4)}^2}{{\sec }^2}\theta (4)\sqrt {{{\sec }^2}\theta - 1} }}} \)

= \(\begin{aligned}{l}\int {\frac{{\tan \theta d\theta }}{{{{(4)}^2}\sec \theta \tan \theta }}} \\\frac{1}{{16}}\int {\frac{1}{{\sec \theta }}d\theta } \\\frac{1}{{16}}\int {\cos \theta d\theta } \\\frac{1}{{16}}\sin \theta + C\end{aligned}\)

=\(\int {\frac{{dt}}{{{t^2}\sqrt {{t^2} - 16} }}} \)=\(\frac{1}{{16}}\sin \theta + c\)

Since, \(t = t\sec \theta \Rightarrow \theta = {\sec ^{ - 1}}\left( {\frac{t}{4}} \right)\)

\( \Rightarrow \int {\frac{{dt}}{{{t^2}\sqrt {{t^2} - 16} }} = \frac{1}{{16}}\sin ({{\sec }^{ - 1}}\left( {\frac{t}{4}} \right))} + C\)

Hence, the indefinite integration of

\( \Rightarrow \int {\frac{{dt}}{{{t^2}\sqrt {{t^2} - 16} }} = \frac{1}{{16}}\sin ({{\sec }^{ - 1}}\left( {\frac{t}{4}} \right))} + C\)