Solve the integral$\int x^3\sqrt{x^2-1}dx$three ways:

(a) with the substitution $u=x^2-1,$followed by back substitution;

(b) with integration by parts, choosing localid="1648814744993" $u=x^2\mathrm{and}dv=x\sqrt{x^2-1}dx;$

(c) with the trigonometric substitution x = sec u.

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

We have to solve the integral with the substitution $u=x^2-1,$so the derivative is$du=2xdx.$

Let's solve the integral by substituting u,

localid="1648816067865" $$\begin{gathered} \int x^3\sqrt{x^2-1}dx \\ =\frac{1}{2}\int \left(u+1\right)\sqrt{u}du \\ =\frac{1}{2}\int u^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}+\sqrt{u}du \\ =\frac{1}{2}\int u^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}du+\frac{1}{2}\int \sqrt{u}du \\ =\frac{1}{2}\left[\frac{2}{5}{u}^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right]+\frac{1}{2}\left[\frac{2}{3}{u}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right]+C \\ =\frac{1}{5}{u}^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}+\frac{1}{3}{u}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}+C \\ \mathrm{Substitute}\mathrm{back}u, \\ =\frac{1}{5}{\left(x^2-1\right)}^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}+\frac{1}{3}{\left(x^2-1\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}+C \\ =\frac{1}{15}\left({\left(x^2-1\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\left(3x^2+2\right)\right)+C \end{gathered}$$

We have to solve the integral with integration by parts, choosing $u=x^2$and $dv=x\sqrt{x^2-1}dx.$

Thus, the derivation of uis $du=\left(2x\right)dx,$ and to find vintegrate dv.

So,

role="math" localid="1648817215834" $$\begin{gathered} \int dv=\int x\sqrt{x^2-1}dx \\ v=\frac{1}{3}{\left(x^2-1\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}dx \end{gathered}$$

Let's do the integration by parts,

$$\begin{gathered} \int x^2\left(x\sqrt{x^2-1}\right)dx \\ =x^2\left(\frac{1}{3}{\left(x^2-1\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right)-\frac{1}{3}\int {\left(x^2-1\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\left(2xdx\right) \\ =\frac{x^2}{3}{\left(x^2-1\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}-\frac{2}{3}\int {\left(x^2-1\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\left(xdx\right) \\ =\frac{x^2}{3}{\left(x^2-1\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}-\frac{2}{3}\left[\frac{1}{5}{\left(x^2-1\right)}^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right]+C \\ =\frac{1}{15}{\left(x^2-1\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\left(5x^2-2\left(x^2-1\right)\right)+c \\ =\frac{1}{15}{\left(x^2-1\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\left(3x^2+2\right)+C \end{gathered}$$

We have to solve the integral with the substitution $x=secu,$so the derivative is$dx=secu\tan udu.$

Let's solve the integral by substituting x,

role="math" localid="1648817837174" $$\begin{gathered} \int x^3\sqrt{x^2-1}dx \\ =\int se{c}^{3}u\sqrt{se{c}^{2}u-1}\left(secu\tan u\right)du \\ =\int se{c}^{4}u\sqrt{{\tan }^{2}u}\left(\tan u\right)du \\ =\int se{c}^{4}u{\tan }^{2}udu \\ =\int se{c}^{2}use{c}^{2}u{\tan }^{2}udu \\ =\int (1+t{\mathrm{an}}^{2}u)se{c}^{2}u{\tan }^{2}udu \\ \mathrm{Let}t=tanu,dt=se{c}^{2}udu \\ =\int (1+t^2)t^{2}dt \\ =\int t^2+t^{4}dt \\ =\int t^{2}dt+\int t^{4}dt \\ =\frac{t^3}{3}+\frac{t^5}{5}+C \end{gathered}$$

By proceeding with the calculation further,

$$\begin{gathered} \mathrm{Substitute}\mathrm{back}t, \\ =\frac{{\left(\mathrm{tanu}\right)}^3}{3}+\frac{{\left(\mathrm{tanu}\right)}^5}{5}+\mathrm{C} \\ =\frac{{\left(\sqrt{{\sec }^2\mathrm{u}-1}\right)}^3}{3}+\frac{{\left(\sqrt{{\sec }^2\mathrm{u}-1}\right)}^5}{5}+\mathrm{C} \\ =\frac{{\left({\sec }^2\mathrm{u}-1\right)}^{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}{3}+\frac{{\left({\sec }^2\mathrm{u}-1\right)}^{{\displaystyle \raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}{5}+\mathrm{C} \\ \mathrm{Substitute}\mathrm{back}u, \\ =\frac{{\left({\mathrm{x}}^2-1\right)}^{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}{3}+\frac{{\left({\mathrm{x}}^2-1\right)}^{{\displaystyle \raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}{5}+\mathrm{C} \\ =\frac{1}{15}\left({\left({\mathrm{x}}^2-1\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\left(3{\mathrm{x}}^2+2\right)\right)+\mathrm{C} \end{gathered}$$