Integration with substitution of trig: $\int\frac1{(x^2-4)^{3/2}}dx$

Question

$$\int\frac1{(x^2-4)^{3/2}}dx $$

I'm unsure how to go about this integral after subbing $x=2\cosh(\theta)$; to the power of $3/2$ is confusing me when trying to simplify.

Thanks.

Answer 1

You may apply $$ \cosh^2 u-\sinh^2 u=1 $$ by setting $$x:=2\cosh u,\qquad (x\geq2),$$ giving $$ (x^2-4)^{3/2}=(4\cosh^2 u-4)^{3/2}=8\left(\sinh^2 u\right)^{3/2}=8\sinh^3 u. $$ Then

$$ \begin{align} \int\frac1{(x^2-4)^{3/2}}dx&=\int\frac{2\sinh u}{8\sinh^3 u}du\\\\ &=\frac14\int\frac{1}{\sinh^2 u}du\\\\ &=-\frac14\coth u+C\\\\ &=-\frac{x}{4 \sqrt{x^2-4}}+C. \end{align} $$

Answer 2

Let $u^2x^2=x^2-4$. Then $x=\pm\sqrt{\frac{-4}{u^2-1}}$, $dx=\pm(-4)\sqrt{\frac{u^2-1}{-16}}\left(\frac{-2u}{\left(u^2-1\right)^2}\right)\,du$.

$$\int\frac{1}{\left(x^2-4\right)^{3/2}}\, dx=\int\frac{\pm(-4)\sqrt{\frac{u^2-1}{-16}}\left(\frac{-2u}{\left(u^2-1\right)^2}\right)\,du}{u^3\cdot \pm\left(\frac{-4}{u^2-1}\right)\sqrt{\frac{-4}{u^2-1}}}$$

$$=\frac{1}{4}\int\frac{1}{u^2}\, du=\frac{x}{-4\sqrt{x^2-4}}+C$$