Find $ \int \frac {\mathrm{d}x}{(4x^2-1)^{3/2}}$ I have trouble using trig sub. After I get that x = 2x+1, should I substitute back into the original problem's $4x^2$ with $(4(2x+1)^2)$?
 A: HINT:
Using Trigonometric substitutions,  set $2x=\sec\theta$
$$\implies4x^2-1=\tan^2\theta$$
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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$\ds{}$
\begin{align}&\overbrace{\color{#66f}{\large\int{\dd x \over \pars{4x^{2} - 1}^{3/2}}}}
^{\ds{\dsc{x} \equiv \dsc{1 \over t}\ \imp\ \dsc{t} \equiv \dsc{1 \over x}}}\ =\
\int{-\,\dd t/t^{2} \over \pars{4/t^{2} - 1}^{3/2}}
=-\int{t\,\dd t \over \pars{4 - t^{2}}^{3/2}}
=-\pars{4 - t^{2}}^{-1/2}
\\[5mm]&=-\pars{4 - {1 \over x^{2}}}^{-1/2}
=\color{#66f}{\large -\,{x \over \root{4x^{2} - 1}}} + \mbox{a constant}
\end{align}
A: $y=2x-1 \implies \dfrac{1}{2}dy=dx$
$\therefore\displaystyle\int\dfrac{dx}{\left(4x^2-1\right)^{\frac{3}{2}}}=\dfrac{1}{2}\displaystyle\int\dfrac{dy}{y^{\frac{3}{2}}\left(y+2\right)^{\frac{3}{2}}}$
$y=2\tan^2\theta \implies dy=4\sec^2\theta\tan\theta\ d\theta$
$\therefore\dfrac{1}{2}\displaystyle\int\dfrac{dy}{y^{\frac{3}{2}}\left(y+2\right)^{\frac{3}{2}}}=\displaystyle\int\dfrac{2\sec^2\theta\tan\theta\ d\theta}{2^3\tan^3\theta\sec^3\theta}=\dfrac{1}{4}\displaystyle\int\csc \theta \cot \theta \ d\theta$
