Question

integral of 1/(sqrt(x^2 - 9))dx evaluated from 5 to 6

Answer 1

Mm, maybe you can use trigonometric identities

Answer 2

\[x=3\sec u~~\Rightarrow~~dx=3\sec u\tan u~du\]
\[\large\begin{align*}\int_5^6\frac{dx}{\sqrt{x^2-9}}&=\int_{\sec^{-1}(5/3)}^{\sec^{-1}(6/3)}\frac{3\sec u\tan u}{\sqrt{(3\sec u)^2-9}}~du\\
&=\int_{\sec^{-1}(5/3)}^{\sec^{-1}2}\frac{3\sec u\tan u}{\sqrt{9\sec^2 u-9}}~du\\
&=\int_{\sec^{-1}(5/3)}^{\pi/3}\frac{\sec u\tan u}{\sqrt{\sec^2 u-1}}~du\\
&=\int_{\sec^{-1}(5/3)}^{\pi/3}\frac{\sec u\tan u}{\sqrt{\tan^2u}}~du\\
&=\int_{\sec^{-1}(5/3)}^{\pi/3}\sec u~du
\end{align*}\]