# 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

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 $n=0$