Question
Let x and a stand for distance. Is $\int\frac{\text{dx}}{\sqrt{\text{a}^2-\text{x}^2}}=\frac{1}{\text{a}}\sin^{-1}\frac{\text{a}}{\text{x}}$ dimensionally correct?
✓
Answer
Dimension of the left side $=\int\frac{\text{dx}}{\sqrt{(\text{a}^2-\text{x}^2})}=\int\frac{\text{L}}{\sqrt{(\text{L}^2-\text{L}^2)}}=[\text{L}^0]$
Dimension of the right side $=\frac{1}{\text{a}}\sin^{-1}\Big(\frac{\text{a}}{\text{x}}\Big)=[\text{L}^{-1}]$
So, the dimension of $\int\frac{\text{dx}}{\sqrt{(\text{a}^2-\text{x}^2)}}\neq\frac{1}{\text{a}}\sin^{-1}\Big(\frac{\text{a}}{\text{x}}\Big)$
So, the equation is dimensionally incorrect.
Dimension of the right side $=\frac{1}{\text{a}}\sin^{-1}\Big(\frac{\text{a}}{\text{x}}\Big)=[\text{L}^{-1}]$
So, the dimension of $\int\frac{\text{dx}}{\sqrt{(\text{a}^2-\text{x}^2)}}\neq\frac{1}{\text{a}}\sin^{-1}\Big(\frac{\text{a}}{\text{x}}\Big)$
So, the equation is dimensionally incorrect.
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