More Substitution Rule — Question 3

Question 3 - Solution

Because the integrand involves $\sqrt{x^2-a^2}$, a hyperbolic substitution is natural.

Let $$x=3\cosh t.$$ Then $$dx=3\sinh t\,dt, \qquad \sqrt{x^2-9}=\sqrt{9\cosh^2 t-9}=3\sinh t.$$

Substitute into the integral: $$\int \frac{x^2}{\sqrt{x^2-9}}\,dx = \int \frac{9\cosh^2 t}{3\sinh t}\cdot 3\sinh t\,dt = \int 9\cosh^2 t\,dt.$$

Use the identity $$\cosh^2 t=\frac{1+\cosh(2t)}{2}.$$

Then $$\int 9\cosh^2 t\,dt = \frac{9}{2}\int \bigl(1+\cosh(2t)\bigr)\,dt = \frac{9}{2}\left(t+\frac{1}{2}\sinh(2t)\right).$$

Simplify: $$\frac{9}{2}t+\frac{9}{4}\sinh(2t).$$

Return to $x$. Since $$x=3\cosh t \quad\Rightarrow\quad t=\operatorname{arcosh}\!\left(\frac{x}{3}\right),$$ and $$\sinh(2t)=2\sinh t\cosh t =2\left(\frac{\sqrt{x^2-9}}{3}\right)\left(\frac{x}{3}\right) =\frac{2x\sqrt{x^2-9}}{9}.$$

Substitute back: $$\boxed{ \frac{9}{2}\operatorname{arcosh}\!\left(\frac{x}{3}\right) +\frac{x}{2}\sqrt{x^2-9} + C }$$