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 $x=a\sinh u\text{oracoshu}$. (These exercises involve hyperbolic functions.)

$\int \frac{x^3}{{\left(x^2-4\right)}^{3/2}}dx$

Short answer

The value of the integral is$\frac{x^2-8}{\sqrt{x^2-4}}+C$.

Step-by-step solution

Step 1. Given information.

The integral is $\int \frac{x^3}{{\left(x^2-4\right)}^{3/2}}dx$.

Step 2. Substitution.

Let $u=x^2$

Therefore,

$du=2xdx$

Step 3. Value of the integral.

The value of the integral is ,

$$\begin{gathered} \int \frac{x^3}{{\left(x^2-4\right)}^{\frac{3}{2}}}dx=\int \frac{u}{2(u-4{)}^{\frac{3}{2}}}du \\ =\frac{1}{2}\int \frac{u}{(u-4{)}^{\frac{3}{2}}}du \\ =\frac{1}{2}\int \frac{2}{v^2}\left(v^2+4\right)dv\left[v=\sqrt{u-4},dv=\frac{1}{2\sqrt{u-4}}du\right] \\ =\int 1dv+\int \frac{4}{v^2}dv \\ =v-\frac{4}{v} \\ =\sqrt{u-4}-\frac{4}{\sqrt{u-4}} \\ =\sqrt{x^2-4}-\frac{4}{\sqrt{x^2-4}} \\ =\frac{x^2-8}{\sqrt{x^2-4}}+C \end{gathered}$$