Integration of 1 Over (x Square Root of x^2-1)

In this tutorial we shall discuss the integration of 1 over x into square root of x^2-1, and this is another important for of integration.

The integration of \frac{1}{{x\sqrt {{x^2} - 1} }} is of the form

\int {\frac{1}{{x\sqrt {{x^2} - 1} }}dx = } {\sec ^{ - 1}}x + c

To prove this formula, consider

\frac{d}{{dx}}\left[ {{{\sec }^{ - 1}}x + c} \right] = \frac{d}{{dx}}{\sec ^{ - 1}}x + \frac{d}{{dx}}c

Using the derivative formula \frac{d}{{dx}}{\sec ^{ - 1}}x = \frac{1}{{x\sqrt {{x^2} - 1} }}, we have

\begin{gathered} \frac{d}{{dx}}\left[ {{{\sec }^{ - 1}}x + c} \right] = \frac{1}{{x\sqrt {{x^2} - 1} }} + 0 \\ \Rightarrow \frac{d}{{dx}}\left[ {{{\sec }^{ - 1}}x + c} \right] = \frac{1}{{x\sqrt {{x^2} - 1} }} \\ \Rightarrow \frac{1}{{x\sqrt {{x^2} - 1} }} = \frac{d}{{dx}}\left[ {{{\sec }^{ - 1}}x + c} \right] \\ \Rightarrow \frac{1}{{x\sqrt {{x^2} - 1} }}dx = d\left[ {{{\sec }^{ - 1}}x + c} \right]\,\,\,\,{\text{ - - - }}\left( {\text{i}} \right) \\ \end{gathered}

Integrating both sides of equation (i) with respect to x, we have

\int {\frac{1}{{x\sqrt {{x^2} - 1} }}dx} = \int {d\left[ {{{\sec }^{ - 1}}x + c} \right]}

As we know that by definition integration is the inverse process of derivative, so we have

\int {\frac{1}{{x\sqrt {{x^2} - 1} }}dx} = {\sec ^{ - 1}}x + c

As we know that derivative of

\frac{d}{{dx}}{\csc ^{ - 1}}x = - \frac{1}{{x\sqrt {{x^2} - 1} }}

This formula can also be written as

\int {\frac{1}{{x\sqrt {{x^2} - 1} }}dx} = - {\csc ^{ - 1}}x + c