# Derivative of $\cos x$

We all know that the integration of $$\cos(x) \ dx$$ is $$\sin(x) + c$$, Since $$\cos(x) \ dx$$ is area, the dimension is unit squared. So is $$\sin(x) + c$$ also having dimension in unit squares? I mean at $$\sin(\frac{π}{2})$$ the value is $$1$$, so is it presented as $$1$$ unit squared?

• Please use mathjax. Regarding your question: The assumption that an an integral gives an area of dimension unit squared is simply wrong. – maxmilgram May 5 at 16:01
• Sorry I'll learn mathjax as soon as possible. However, isn't it weird how an area doesn't have a dimension of L squared? – NightKruger May 5 at 16:11
• No it isnt, it is not necessarily a spatial area. E.g. if you integrate a velocity over time, you get a length. – maxmilgram May 5 at 18:14

The argument of a trigonometric function is dimensionless. While $$x$$ would have the dimension of $$k^{-1}$$ in $$\cos kx$$, the problem at hand has $$k=1$$ so $$x$$ is dimensionless. We therefore can't uniquely specify a dimension for a trigonometric function, whether we obtained it by differentiation, integration, neither, or one repeated. Note the value of $$\cos kx$$ is dimensionless too, regardless of $$k$$.

• I understand , but isn't it weird how an area doesn't have a dimension L^2? – NightKruger May 5 at 16:12
• @NightcoRohak However, an area under $L_1\cos\frac{x}{L_2}$ with $L_1,\,L_2$ each of dimension $L$ will have dimension $L^2$. – J.G. May 5 at 16:14
• Well that true. So you are saying that the result of an integration is dimensionless if our graph is dimensionless. So the integration of cosxdx is just a number, which happens to equal the length of the graph of sinx + c right? Or am I again missing something. Sorry for the silly questions , but I think you can understand that I'm very new to calculus. – NightKruger May 5 at 16:20
• @NightcoRohak Since the output of $\cos x$ is a dimensionless number, it must be followed by some kind of unit to be able to represent a dimensional physical quantity. In other words, $\cos xdx$ is the magnitude of the area, but the actual area of the element is $\cos xdx$ squared units. Upon integration, the unit/suffixed 'squared units' persists, so the area is actually $(\sin x+c)$ squared units, which further implies that $\sin x$, like its argument $x$, is dimensionless and just gives the magnitude. – Shubham Johri May 5 at 16:25
• @NightcoRohak We were all new to it once. The important thing to understand is that while integration can obtain an area, it's not limited to that. The dimension of an integral (be it definite it but indefinite), with respect to $x$, of $f(x)$ is the dimension of $xf(x)$. It can be anything, especially in physical applications. – J.G. May 5 at 16:27