However, $a x=\sin \theta$
$\Rightarrow \sin \theta \in\left(-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)$
$\Rightarrow \theta \in\left(-\frac{\pi}{4}, \frac{\pi}{4}\right)$
$\Rightarrow 2 \theta \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$
Hence, $u=\sin ^{-1}(\sin 2 \theta)=2 \theta$.
$\Rightarrow u=2 \sin ^{-1}(a x)$
On differentiating $u$ with respect to $x$, we get
$\frac{d u}{d x}=\frac{d}{d x}\left(2 \sin ^{-1} a x\right)$
$\Rightarrow \frac{\mathrm{du}}{\mathrm{dx}}=2 \frac{\mathrm{d}}{\mathrm{dx}}\left(\sin ^{-1} \mathrm{ax}\right)$
We know $\frac{d}{d x}\left(\sin ^{-1} x\right)=\frac{1}{\sqrt{1-x^{2}}}$
$\Rightarrow \frac{d u}{d x}=2 \times \frac{1}{\sqrt{1-(a x)^{2}}} \frac{d}{d x}(a x)$
$\Rightarrow \frac{d u}{d x}=\frac{2}{\sqrt{1-a^{2} x^{2}}}\left[a \frac{d}{d x}(x)\right]$
$\Rightarrow \frac{d u}{d x}=\frac{2 a}{\sqrt{1-a^{2} x^{2}}} \frac{d}{d x}(x)$
We know $\frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{x})=1$
$\Rightarrow \frac{\mathrm{du}}{\mathrm{dx}}=2 \times \frac{1}{\sqrt{1-(\mathrm{ax})^{2}}} \frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{ax})$
$\Rightarrow \frac{\mathrm{du}}{\mathrm{dx}}=\frac{2}{\sqrt{1-\mathrm{a}^{2} \mathrm{x}^{2}}}\left[\mathrm{a} \frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{x})\right]$
$\Rightarrow \frac{\mathrm{du}}{\mathrm{dx}}=\frac{2 \mathrm{a}}{\sqrt{1-\mathrm{a}^{2} \mathrm{x}^{2}}} \frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{x})$
We know $\frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{x})=1$
$\Rightarrow \frac{\mathrm{du}}{\mathrm{dx}}=\frac{2 \mathrm{a}}{\sqrt{1-\mathrm{a}^{2} \mathrm{x}^{2}}} \times 1$
$\therefore \frac{\mathrm{du}}{\mathrm{dx}}=\frac{2 \mathrm{a}}{\sqrt{1-\mathrm{a}^{2} \mathrm{x}^{2}}}$
Now, we have $\mathrm{v}=\sqrt{1-\mathrm{a}^{2} \mathrm{x}^{2}}$
On differentiating $v$ with respect to $x$, we get
$\frac{d v}{d x}=\frac{d}{d x}\left(\sqrt{1-a^{2} x^{2}}\right)$
$\Rightarrow \frac{d v}{d x}=\frac{d}{d x}\left(1-a^{2} x^{2}\right)^{\frac{1}{2}}$
We know $\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{x}^{\mathrm{n}}\right)=\mathrm{nx}^{\mathrm{n}-1}$
$\Rightarrow \frac{d v}{d x}=\frac{1}{2}\left(1-a^{2} x^{2}\right)^{\frac{1}{2}-1} \frac{d}{d x}\left(1-a^{2} x^{2}\right)$
$\Rightarrow \frac{d v}{d x}=\frac{1}{2}\left(1-a^{2} x^{2}\right)^{-\frac{1}{2}}\left[\frac{d}{d x}(1)-\frac{d}{d x}\left(a^{2} x^{2}\right)\right]$
$\Rightarrow \frac{d v}{d x}=\frac{1}{2 \sqrt{1-a^{2} x^{2}}}\left[\frac{d}{d x}(1)-a^{2} \frac{d}{d x}\left(x^{2}\right)\right]$
We know $\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{x}^{\mathrm{n}}\right)=\mathrm{n} \mathrm{x}^{\mathrm{n}-1}$ and derivative of a constant is 0 .
$\Rightarrow \frac{\mathrm{dv}}{\mathrm{dx}}=\frac{1}{2 \sqrt{1-\mathrm{a}^{2} \mathrm{x}^{2}}}\left[0-\mathrm{a}^{2}\left(2 \mathrm{x}^{2-1}\right)\right]$
$\Rightarrow \frac{\mathrm{dv}}{\mathrm{dx}}=\frac{1}{2 \sqrt{1-\mathrm{a}^{2} \mathrm{x}^{2}}}\left[-2 \mathrm{a}^{2} \mathrm{x}\right]$
$\therefore \frac{\mathrm{dv}}{\mathrm{dx}}=-\frac{\mathrm{a}^{2} \mathrm{x}}{\sqrt{1-\mathrm{a}^{2} \mathrm{x}^{2}}}$
We have $\frac{d u}{d v}=\frac{\frac{d u}{d v}}{\frac{d v}{d x}}$
$\Rightarrow \frac{d u}{d v}=\frac{\frac{2 a}{\sqrt{1-a^{2} x^{2}}}}{-\frac{a^{2} x}{\sqrt{1-a^{2} x^{2}}}}$
$\Rightarrow \frac{d u}{d v}=\frac{2 a}{\sqrt{1-a^{2} x^{2}}} \times \frac{\sqrt{1-a^{2} x^{2}}}{-a^{2} x}$
$\therefore \frac{\mathrm{du}}{\mathrm{dv}}=-\frac{2}{\mathrm{ax}}$
Thus, $\frac{\mathrm{du}}{\mathrm{dv}}=-\frac{2}{\mathrm{ax}}$
