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Help please?

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, 7 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, \( [L^2] - [L^2] = [L^2] \)

Siddhartha Srivastava - 2 years, 7 months ago

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Oh okay! I overlooked that. Thanks! :D

Omkar Kulkarni - 2 years, 7 months ago

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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 \(x=\)\( a\sin { \theta }\)

Soumya Dubey - 2 years, 7 months ago

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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 \(x=a\sin\theta\)?

Omkar Kulkarni - 2 years, 7 months ago

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