Evaluate $\displaystyle\int \dfrac{x}{\left(16 - 9 x^2\right)^{3/2}} \; dx$ $\;\;$ $\left(0 < x < \dfrac{4}{3}\right)$
$\begin{aligned}
\text{Let } I & = \int \dfrac{x}{\left(16 - 9 x^2\right)^{3/2}} \; dx \\\\
& = \int \dfrac{x}{\left\{\sqrt{9 \left(\dfrac{16}{9} - x^2\right)} \right\}^3} \; dx \\\\
& = \dfrac{1}{27} \int \dfrac{x \; dx}{\left\{\sqrt{\left(\dfrac{4}{3}\right)^2 - \left(x\right)^2} \right\}^3} \;\;\; \cdots (1)
\end{aligned}$
Put $x = \dfrac{4}{3} \sin \theta$ $\;\;\; \cdots$ (2a)
Then, $dx = \dfrac{4}{3} \cos \theta \; d \theta$ $\;\;\; \cdots$ (2b)
In view of equations (2a) and (2b), equation (1) can be written as
$\begin{aligned}
I & = \dfrac{1}{27} \int \dfrac{\dfrac{4}{3} \; \sin \theta \times \dfrac{4}{3} \; \cos \theta \; d \theta}{\left\{\sqrt{\left(\dfrac{4}{3}\right)^2 - \left(\dfrac{4}{3} \sin \theta\right)^2} \right\}^3} \\\\
& = \dfrac{1}{27} \times \dfrac{16}{9} \int \dfrac{\sin \theta \; \cos \theta \; d \theta}{\left\{\sqrt{\dfrac{16}{9} \left(1 - \sin^2 \theta\right)} \right\}^3} \\\\
& = \dfrac{1}{27} \times \dfrac{16}{9} \int \dfrac{\sin \theta \; \cos \theta \; d \theta}{\left(\dfrac{4}{3} \; \cos \theta\right)^3} \\\\
& = \dfrac{1}{27} \times \dfrac{16}{9} \times \dfrac{27}{64} \int \dfrac{\sin \theta \; \cos \theta \; d \theta}{\cos^3 \theta} \\\\
& = \dfrac{1}{36} \int \dfrac{\sin \theta \; d \theta}{\cos^2 \theta} \\\\
& = \dfrac{1}{36} \int \dfrac{1}{\cos \theta} \times \dfrac{\sin \theta}{\cos \theta} \; d \theta \\\\
& = \dfrac{1}{36} \int \sec \theta \; \tan \theta \; d \theta \\\\
& = \dfrac{1}{36} \sec \theta + c \;\;\; \cdots (3)
\end{aligned}$
From equation (2a),
$\sin \theta = \dfrac{3 x}{4}$
$\therefore$ $\;$ $\cos \theta = \sqrt{1 - \sin^2 \theta} = \sqrt{1 - \dfrac{9 x^2}{16}} = \dfrac{\sqrt{16 - 9 x^2}}{4}$
$\therefore$ $\sec \theta = \dfrac{4}{\sqrt{16 - 9x^2}}$ $\;\;\; \cdots$ (4)
Substituting the value of $\sec \theta$ from equation (4) in equation (3) gives
$\begin{aligned}
I & = \dfrac{1}{36} \times \dfrac{4}{\sqrt{16 - 9 x^2}} + c \\\\
& = \dfrac{1}{9} \times \dfrac{1}{\sqrt{16 - 9 x^2}} + c
\end{aligned}$