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?
 A: Let
$$
I=\int \frac{1}{\left(1 + cx^2\right)^\frac{3}{2}} \mathrm{d}x= \int \frac{1}{x^3(1/x^2+c)^{3/2}}$$.
Let $1/x=u \implies -dx/x^2=du$ and then $u^2=v$
$$I=-\int \frac{udu}{(u^2+c)^{3/2}}=\frac{-1}{2} \int (v+c)^{-3/2} dv=(v+c)^{-1/2}.$$
Finally $$I=\frac{x}{\sqrt{1+cx^2}}+C$$
A: Here's one idea that requires $c > 0$ and $a,b \in \mathbb{R}$. Let $I$ be the given integral. If $x = \frac{\sinh{(u)}}{\sqrt{c}}$, then the integral becomes
$$
\eqalign{
I &= \frac{1}{\sqrt{c}}\int_{\operatorname{arcsinh}\left(a\sqrt{c}\right)}^{\operatorname{arcsinh}\left(b\sqrt{c}\right)}\operatorname{sech}^{2}\left(u\right)du \cr
 &= \frac{1}{\sqrt{c}}\tanh{(u)}\Bigg|_{\operatorname{arcsinh}\left(a\sqrt{c}\right)}^{\operatorname{arcsinh}\left(b\sqrt{c}\right)} \cr
&= \frac{1}{\sqrt{c}}\left(\tanh\left(\operatorname{arcsinh}\left(b\sqrt{c}\right)\right)-\tanh\left(\operatorname{arcsinh}\left(a\sqrt{c}\right)\right)\right) \cr
&= \frac{b}{\sqrt{cb^{2}+1}}-\frac{a}{\sqrt{ca^{2}+1}}.
}
$$
A: let ${(1+cx^2)}^{-1/2}=u \rightarrow 1 $
so you have $du= \frac{-1}{2} \frac{2x}{(1+cx^2)^{-3/2}}dx$
from 1
$x=\sqrt{\frac{1}{u^2} -c}$
so the new integrand  is $\frac{{-du}}{\sqrt{\frac{1}{u^2} -c}}$
which is the same as
$$\frac{-udu}{\sqrt{1-cu^2}}$$
I think it is doable from here
A: $$
\begin{aligned}
&\text { Let } y=\frac{1}{x^2} \text {, then } d x=-\frac{1}{2y^{\frac{3}{2}}} d y\\
I&=\int_{\frac{1}{a^2}}^{\frac{1}{b^2}} \frac{1}{\left(1+\frac{c}{y}\right)^{\frac{3}{2}}}\left(-\frac{d y}{2y \frac{3}{2}}\right)\\
&=\frac{1}{2} \int_{\frac{1}{b^2}}^{\frac{1}{a^2}} \frac{d y}{(y+c)^{\frac{3}{2}}}\\
&=\left[-\frac{1}{\sqrt{y+c}}\right]_{\frac{1}{b^2}}^{\frac{1}{a^2}}\\
&=\frac{b}{\sqrt{1+b^2 c}}-\frac{a}{\sqrt{1+a^2 c}}
\end{aligned}
$$
