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Can we apply dimensional analysis for variables inside integrals? Ex: if we have integral $$\int \frac{\text{d}x}{\sqrt{a^2 - x^2}} = \frac{1}{a} \sin^{-1} \left(\frac{a}{x}\right),$$ the LHS has no dimensions, while the RHS has dimensions of $\frac{1}{length}$. So let me know , whether I am correct?

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Yes, you can apply dimensional analysis to integrals. You count differentials like $dx$ as having the units of the associated variable, because $dx$ can be interpreted as an infinitesimal change in $x$.

In your example, if $a$ and $x$ have units of distance, then checking the units in your result shows that it's incorrect. The correct integral does not have the factor of $1/a$.

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To be more explicit about Ben Crowell's remarks: Let $u = \frac xa$, so $dx = adu$ and the integration becomes $$\int \frac{dx}{\sqrt{a^2-x^2}} = \int \frac{adu}{\sqrt{a^2-a^2u^2}} = \int \frac{du}{\sqrt{1 - u^2}} = \sin^{-1} u + C = \sin^{-1}\frac xa + C$$ (assuming that $a> 0$).

Since $a$ and $x$ must have the same units for the expression $a^2 - x^2$ to make sense, we see that $u$ is dimensionless, as is the integrand in all 3 versions, and as is the correct integral.

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Yes you may. Think of an integral $\int g(x) dx$ as the sum $\sum g(x) \Delta x$. In each term in your (very large) summation, you may apply dimensional analysis.

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Your integral has a symmetry: if you map $a \mapsto k a$ and then $x \mapsto kx$ your RHS reads:

$$ \sin^{-1} \frac{a}{x} \mapsto \sin^{-1} \frac{ka}{kx} = \sin^{-1} \frac{a}{x} $$

That went well... what about the left hand side? Does was that also invariant?

$$ \int \frac{dx}{\sqrt{a^2 -x^2}} \mapsto \int \frac{d(kx)}{\sqrt{(ka)^2 -(kx)^2}} = \int \frac{k\,dx}{k\sqrt{a^2 -x^2}} = \int \frac{dx}{\sqrt{a^2 -x^2}}$$ So - by the rules of calculus - these integrals are both fixed points of the rescaling action $(a,x) \mapsto (ka,kx)$. We really needed the linearity of the integral and differential here $\int$ and $d$.

There may even be other symmetries. For one thing, RHS is not unless - it can be thought of as an angle $\theta$ with units of radians. What happens when we change the integration constant $\theta \mapsto \theta + c $ ?

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  • $\begingroup$ Angles are not unitless, but they are dimensionless in the usual dimension systems. In some calculation, you can distinguish “parallel length” and “perpendicular length” to get angles with dimension of $\frac{perpendicular\ length}{parallel\ length}$, but you can't do that generally. $\endgroup$ – Jan Hudec Jun 9 '18 at 10:16

protected by Qmechanic Jun 7 '18 at 17:15

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