Question

Let x and a stand for distance. Is $\int \frac{dx}{\sqrt{a^2-x^2}}=\frac{1}{a}{\sin }^{-1}\frac{a}{x}$ dimensionally correct?

Answer 1

Dimension of the left side of the equation= $\left[\int \frac{dx}{\sqrt{\left(a^2-x^2\right)}}\right]=\left[\int \frac{\mathrm{L}}{\sqrt{\left({\mathrm{L}}^2-{\mathrm{L}}^2\right)}}\right]=\left[{\mathrm{L}}^0\right]$

Dimension of the right side of the equation = $\left[\left(\frac{1}{a}\right){\sin }^{-1}\left(\frac{a}{x}\right)\right]=\left({\mathrm{L}}^{-1}\right)$

So, $\left[\int \frac{dx}{\sqrt{\left(a^2-x^2\right)}}\right]\ne \left[\left(\frac{1}{a}\right){\sin }^{-1}\left(\frac{a}{x}\right)\right]$

Since the dimensions on both sides are not the same, the equation is dimensionally incorrect.