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प्रश्न
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?
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उत्तर
Dimension of the left side of the equation= \[{\left[ \int\frac{dx}{\sqrt{\left( a^2 - x^2 \right)}} \right]} = {\left[ \int\frac{L}{\sqrt{\left( L^2 - L^2 \right)}} \right]} = \left[ L^0 \right]
\text{Dimension of the right side of the equation} = \left[ \left( \frac{1}{a} \right) \sin^{- 1} \left( \frac{a}{x} \right) \right] = \left( L^{- 1} \right)\]
So,
\[ {\left[ \int\frac{dx}{\sqrt{\left( a^2 - x^2 \right)}} \right]} \neq \left[ \left( \frac{1}{a} \right) \sin^{- 1} \left( \frac{a}{x} \right) \right]\]
Since the dimensions on both sides are not the same, the equation is dimensionally incorrect.
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