Understanding change of variables

Question

So there's this one thing I can't wrap my head around.

The problem is:

Use the change of variables $x = \sinh u$ to compute the indefinite integral \begin{equation*} \int \frac{dx}{\sqrt{1+x^2}} \end{equation*}

Solution:

\begin{align*} \frac{dx}{du}=\sinh'x=\cosh x \end{align*} $$\iff$$ \begin{align*} \int \frac{\cosh x}{\sqrt{1+\sinh^2x}}\,du = \int \frac{\cosh x}{\sqrt{\cosh^2x}}\,du = \int\frac{\cosh x}{\cosh x}\,du= \int 1 \,du = u + C = \sinh^{-1}x+C \end{align*}

and I understand every step except the first one, $\frac{dx}{du}=\sinh'x$. I'm just generally confused when it comes to change of variables and how to use the Leibniz notation to understand what's going on.

Answer 1

I think it's clearer if you use differentials: remember the differential of a function $f$ of the variable $u$ is defined as the formal expression $$\mathrm df=f'(u)\,\mathrm du, $$ where $\mathrm du$ (the differential of the variable) represents symbolically a small increment of the variable, and $\mathrm df$ (the differential of the function) is an approximation of the corresponding increment of $f(u)$, as measured along the tangent line.

This being said, the substitution $x=\sinh u$ yields the differential of $x$: $$\mathrm dx=\cosh u\,\mathrm du, $$ and we have to express the whole differential form $\frac{\mathrm dx}{\sqrt{1+x^2}}$ in function of $u$ and its differential. Therefore, the integral becomes $$\int\frac{\mathrm dx}{\sqrt{1+x^2}}\to \int\frac{\cosh u\,\mathrm du}{\sqrt{1+\sinh^2u}}=\int\frac{\cosh u\,\mathrm du}{\cosh u}=\int\mathrm du=u=\operatorname{argsinh}x.$$

Answer 2

Note that if $x=\sinh(u)$, so you have $$\frac{dx}{du}=\frac{d}{du}\left(\sinh(u) \right)=\cosh(u)$$ It's the same that you write $$x=\sinh(u) \implies x'=\cosh(u)$$ However, the second notation it's not recommended for when you are integrating because this may be confused you.