Question

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

Answer 1

Dimension of the left side

$\int \frac{dx}{\sqrt{(a^2-x^2)}}=\int \frac{L}{\sqrt{(L^2-L^2)}}=\left[L^{\circ }\right]$

Dimension of the right side

$\left(\frac{1}{a}\right)si{n}^{-1}\left(\frac{a}{x}\right)=\left(L^{-1}\right)$

So, dimension of

$\int \frac{dx}{\sqrt{(a^2-x^2)}}\ne \left(\frac{1}{a}\right)si{n}^{-1}\left(\frac{a}{x}\right)$

Thus the equation is dimensionally incorrect.