# Can you explain this equation to someone who hasn't learned integration?

So I have just entered 11th grade and started limits on my own but my Physics textbook has an equation which I don't understand, I suspect it uses integration which I haven't learned yet. So can someone explain this equation to me:-

The question is:-

Find the value of $n$ (by using dimensional analysis): $$\int \frac{dx}{\sqrt{2ax-x^2}} = a^n \sin^{-1} [\frac{x}{a} -1]$$

The equation looks similar to $$v^2-u^2 = 2ax$$ But I don't understand what $dx$ means.

• There's no way around learning what an integral means here. The left-hand side of the equation is an integral, and so the exercise can't make sense to you until you know integrals. – Henning Makholm May 24 '15 at 12:21

## 2 Answers

There's not really an alternative to learning what an integral means here. The left-hand side of the equation is an integral, and so the exercise can't make sense to you until you know integrals.

The best we can do without writing an entire textbook here is:

For the purpose of dimensional analysis it is enough to know that an integral $\int f(x)\,dx$ is a limit of certain sums of terms of the form $f(x_1)\cdot(x_2-x_3)$. So your left-hand side $$\int \frac{dx}{\sqrt{2ax-x^2}}$$ has the same dimension as $$\frac{x_2-x_3}{\sqrt{2ax_1-x_1^2}}$$ where of course all the $x_i$s have the same dimension as the $x$. This turns out to mean that the value of the integral is dimensionless.

Let's try dimensional analysis the LHS is a sum of quantities that have $[x]-\sqrt{[x]^2}$ dimension i.e the integral is dimensionless.

The RHS should be dimensionless. The arcsine is dimensionless $a$ has the dimension of $[x]$ (in the denominator of the integral $ax$ is added to $x^2$) so $n=0$