Integration of $\sin^{-1} (x)$ using Lagrange notation I would like to find the anti-derivative of $\sin^{-1} (x)$ using Lagrange notation.
For derivation, Lagrange uses $f'(x)$, $f''(x)$, etc. However, for anti-derivation he uses $\int$ (elongated $S$) symbol and $f(x)$ in it.
We would like to compute the derivative of $\sin^{-1} (x)$. Let $y = \sin^{-1} (x)$. Here $y$ is the function of $x$, so $f(x) = \sin^{-1}  (x)$. Now, $\sin (y) = x$. then, we can also write it as $\sin [f(x)] = x$, since $y = f(x)$.
Then, differentiating using the chain rule,
$$\cos[f(x)] f'(x) = 1.$$
$$f'(x) = 1/\cos[f(x)].$$
Using trigonometry, $$\sin^2 [f(x)] + \cos^2 [f(x)] = 1,$$ $$\cos[f(x)] = \sqrt{ 1 - \sin^2[f(x)] }.$$ Which is equivalent to $\sqrt{1 - x^2}$.
So $f'(x) = 1 / \sqrt{1 - x^2}$.
Here $f'(x)$ means the derivative of function of $x$ w.r.t $x$, and $y$ is a function of $x$.
So by Leibniz notation, we can also write $dy/dx = f'(x)$, here $dy/dx$ also means the derivative of a function of $x$ w.r.t $x$ since $y$ is a function of $x$.
Similarly, how can I find the anti-derivative of $\sin^{-1} (x)$ using Lagrange notation, using as few symbols as possible; I don't want to treat $dy/dx$ like a ratio.
 A: The following method is equivalent to making the substitution $u=\arcsin(x)$, but it avoids treating $du/dx$ as a ratio. Notice that
$$
\int \arcsin(x) \, dx = \int \color{red}{\arcsin(x) \cdot \sqrt{1-\sin^2(\arcsin(x))}} \cdot \color{blue}{\frac{1}{\sqrt{1-x^2}}} \, dx \, .
$$
This integral is of the form
$$
\int \color{red}{f'(g(x))} \cdot \color{blue}{g'(x)} \, dx \, ,
$$
with $f'(u)=u\cdot\sqrt{1-\sin^2(u)}$ and $g(x)=\arcsin(x)$. Since
$$
\int f'(g(x)) \cdot g'(x) \, dx =f(g(x))+C \, ,
$$
we get that
$$
\int \arcsin(x) \cdot \sqrt{1-\sin^2(\arcsin(x))} \cdot \frac{1}{\sqrt{1-x^2}} \, dx = f(\arcsin (x)) \, ,
$$
where $f$ is an antiderivative of $u \cdot \sqrt{1-\sin^2(u)}$. So the problem reduces to integrating $u\cdot\sqrt{1-\sin^2(u)}$, which we can doing using integration by parts:
\begin{align}
\int u \cdot \sqrt{1-\sin^2(u)} &= \int u \cos(u) \, du \\[4pt]
&= u\sin(u)-\int \sin(u) \, du\\[4pt]
&= u\sin(u)+\cos(u) + C \, .
\end{align}
Hence, $f(u)=u\sin(u)+\cos(u)+C$, and so
\begin{align}
\int \arcsin(x) \, dx &= \arcsin(x)\sin(\arcsin(x)) + \cos(\arcsin(x)) + C \\[4pt]
&= x\arcsin(x) + \sqrt{1-x^2} + C \, .
\end{align}
