Question

We can extend the technique of trigonometric substitution to the hyperbolic functions. Use Theorem 2.20 and the identity ${\cosh }^{2}x-{\sinh }^{2}x=1$to solve each integral in Exercises 87– 90 with an appropriate hyperbolic substitution $x=a\sinh u\mathrm{or}x=a\cos hu$. (These exercises involve hyperbolic functions.)

$\int \frac{1}{\sqrt{2x^2-3}}dx$

Short answer

The value of the integral is$\frac{\sqrt{2}}{2}{\cosh }^{-1}\left(\frac{\sqrt{6}x}{3}\right)+C$.

Step-by-step solution

Step 1. Given information.

The given integral is$\int \frac{1}{\sqrt{2x^2-3}}dx$.

Step 2. Substitution.

Now, Let $x=\frac{\sqrt{6}}{2}\cosh \left(u\right)$

Then,

$$\begin{gathered} x=\frac{\sqrt{6}}{2}\cosh \left(u\right) \\ dx=\frac{\sqrt{6}}{2}\sinh \left(u\right)du \end{gathered}$$

Since ${\cosh }^2\left(u\right)-1={\sinh }^2\left(u\right),$

$$\begin{gathered} \frac{1}{\sqrt{2x^2-3}}=\frac{1}{\sqrt{3{\cosh }^2\left(u\right)-3}} \\ =\frac{\sqrt{3}}{3\sqrt{{\cosh }^2\left(u\right)-1}} \\ =\frac{\sqrt{3}}{3\sqrt{{\sinh }^2\left(u\right)}} \\ =\frac{\sqrt{3}}{3\sinh \left(u\right)} \end{gathered}$$

Step 3. Value of the integral.

Now, we can write,

$$\begin{gathered} \int \frac{1}{\sqrt{2x^2-3}}dx=\int \frac{\sqrt{2}}{2}du \\ =\frac{\sqrt{2}}{2}u \\ =\frac{\sqrt{2}}{2}{\cosh }^{-1}\left(\frac{\sqrt{6}x}{3}\right)+C \end{gathered}$$