Integrating a time derivative

Question

let $x$ be a function of time $$x = x(t)$$ what is the integral of $$\int \frac{\dot x}{ \sqrt{1+ \dot x^2}} \, dt$$

I have tried using trigonometric substitution of $$ \dot x = \tan(\theta) $$

but I end up at the same point but with trigonometric terms

Answer 1

You can integrate by parts,

$$\int \frac{\dot x}{\sqrt{1+\dot x^2}}\mathrm d x=\frac{x}{\sqrt{1+\dot x^2}}+\int \frac{\dot x^2 \ddot x}{(1+\dot x ^2)^{\frac32}} \mathrm d x.$$

Since $\mathrm{d} \dot x = \ddot x \; \mathrm{d} t$, plug into the previous integral,

$$\int \frac{\dot x}{\sqrt{1+\dot x^2}}\mathrm d x=\frac{x}{\sqrt{1+\dot x^2}}+\int \frac{\dot x^2}{(1+\dot x ^2)^{\frac32}} \mathrm d \dot x,$$

which is solvable by the trigonometric substitution you proposed.