Is dimensional analysis valid for integrals 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?
 A: 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 $1/a$.
A: 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 $ ?
A: 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.
A: 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.
