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I came across this question in a book.

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?

And this is a solution I found.

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

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

So, the dimension of $$\int\frac{dx}{\sqrt{a^{2}-x^{2}}}\neq\frac{1}{a}\sin^{-1}\frac{a}{x}$$

So, the equation is dimensionally incorrect.

Could someone explain me how $$\int\frac{L}{\sqrt{L^{2}-L^{2}}}=[L^{0}]$$? Shouldn't the denominator be zero or something?

Note by Omkar Kulkarni
2 years, 2 months ago

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Remember we're not talking about numbers anymore. We're talking about dimensions. Rules of algebra do not nessecarily apply here. So $$[L^2] - [L^2]$$ is not nessecarily $$0$$. Both $$x^2$$ and $$a^2$$ have a dimesnion of area. So when we add them or subtact them, the resulting quantity will also have the same dimension of area. (as long as the sum/differnce is not 0 ).

Look at it this way. If two sqaures have an area of $$9 m^2$$ and $$4 m^2$$, what is the difference of the areas? $$5m^2$$. The equation here is $$9m^2 - 4m^2 = 5m^2$$. Or, if you look at this dimensionally, $$[L^2] - [L^2] = [L^2]$$ · 2 years, 2 months ago

Oh okay! I overlooked that. Thanks! :D · 2 years, 2 months ago

i think the answer to your integral is wrong ,the given integral is :-

$$\int { \frac { dx }{ \sqrt { { a }^{ 2 }-{ x }^{ 2 } } } } =\arcsin { \frac { x }{ a } } +k$$

now recheck the dimensions.

u can recheck for yourself ,just substitute $$x=$$$$a\sin { \theta }$$ · 2 years, 2 months ago

Yeah, there's a similar question which has $$\sin^{-1}\left[\frac{x}{a}+1\right]$$. But this integral is meant to be wrong. Thanks though! But why can you substitute $$x=a\sin\theta$$? · 2 years, 2 months ago