Is a trig substitution the only way to solve $ \int_a^b \frac{1}{\left(1 + cx^2\right)^{3/2}} \mathrm{d}x $? I have an integral that looks like the following:
$$
\int_a^b \frac{1}{\left(1 + cx^2\right)^{3/2}} \mathrm{d}x
$$
I have seen a method of solving it being to substitute $x = \frac{\mathrm{tan}(u)}{\sqrt{c}}$; however, this seems somewhat sloppy to me. Is there perhaps a better way of tackling this integral?

$$
\begin{aligned}
&\text { Let } y=\frac{1}{x^2} \text {, then } d x=-\frac{1}{2y^{\frac{3}{2}}} d y\\
I&=\int_{\frac{1}{a^2}}^{\frac{1}{b^2}} \frac{1}{\left(1+\frac{c}{y}\right)^{\frac{3}{2}}}\left(-\frac{d y}{2y \frac{3}{2}}\right)\\
&=\frac{1}{2} \int_{\frac{1}{b^2}}^{\frac{1}{a^2}} \frac{d y}{(y+c)^{\frac{3}{2}}}\\
&=\left[-\frac{1}{\sqrt{y+c}}\right]_{\frac{1}{b^2}}^{\frac{1}{a^2}}\\
&=\frac{b}{\sqrt{1+b^2 c}}-\frac{a}{\sqrt{1+a^2 c}}
\end{aligned}
$$
