What does this equation mean? [closed]

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.

closed as off-topic by Kyle Kanos, ACuriousMind♦, Rob Jeffries, Martin, DanuMay 26 '15 at 9:55

• This question does not appear to be about physics within the scope defined in the help center.
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• dx is just the variable with which to integrate with respect to. like $\frac{dy}{dx}$ is differentiating y with respect to x. Here, you integrate with respect to x also. – Weasel May 24 '15 at 11:50
• I think, this is rather a question for math.SE, as it is concerded only with mathematical concepts. – Sebastian Riese May 24 '15 at 12:00
• It would help you would just give a general understanding, you don't have to go too deep. – Abhishek Mhatre May 24 '15 at 12:05
• Perhaps you should ask the question here : artofproblemsolving.com/community/… – user37026 May 24 '15 at 12:49
• It's a calculus formula. This is not the right place to learn about how calculus works. – Rob Jeffries May 25 '15 at 9:54

Clearly $a$ has the same dimension of $x$ (see the argument of root or of $\sin^{-1}$) so the left member is dimensionless (ratio between dimension of x: remember that differential dx count in dimensional calculus!), and the second member too has to be dimensionless: so n=0.
Integration is finding the area under a curve that isn't necessarily straight. If you have a velocity time graph and find the area under it, this gives you the distance travailed. If you have a acceleration-time graph the area under it is the change in velocity. There are several techniques to integration, which I will not go into here. As mentioned in the comments $dx$ tells you, you are integrating w.r.t. $x$ it is a label rather then a physical quantity in this sense. When doing dimensional analysis you can simply give the dx a dimension of length then and ignore the integral sign $\int$.