Let x and a stand for distance- Is -1-a2-x2dx-1-a sin-1a-x dimensionally correct -

Prove the statement:Dimension of the left side∫1/√((a^2 x^2))dx = [L^0]Dimension of the right side 1/a sin^ 1(a/x) = [L^ 1]Since the dimension of L.H.S. dime ...

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Standard XII

Mathematics

Property 1

Let x and a s...

Question

Let $x$ and $a$ stand for distance. Is $\int \frac{1}{\sqrt{(a^{2} - x^{2})}} d x = \frac{1}{a} \sin^{- 1} (\frac{a}{x})$ dimensionally correct ?

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Solution

Prove the statement:
Dimension of the left side $\int \frac{1}{\sqrt{(a^{2} - x^{2})}} d x = [L^{0}]$
Dimension of the right side $\frac{1}{a} \sin^{- 1} (\frac{a}{x}) = [L^{- 1}]$
Since the dimension of L.H.S. $\neq$ dimension of R.H.S.
Hence, the given equation is dimensionally incorrect.

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Let x and a stand for distance. Is ∫1(a2-x2)dx=1asin-1ax dimensionally correct ?

Prove the statement:Dimension of the left side∫1(a2-x2)dx=[L0]Dimension of the right side 1asin-1ax=[L-1]Since the dimension of L.H.S. ≠dimension of R.H.S.Hence, the given equation is dimensionally incorrect.

Let $x$ and $a$ stand for distance. Is $\int \frac{1}{\sqrt{(a^{2} - x^{2})}} d x = \frac{1}{a} \sin^{- 1} (\frac{a}{x})$ dimensionally correct ?

Prove the statement:
Dimension of the left side $\int \frac{1}{\sqrt{(a^{2} - x^{2})}} d x = [L^{0}]$
Dimension of the right side $\frac{1}{a} \sin^{- 1} (\frac{a}{x}) = [L^{- 1}]$
Since the dimension of L.H.S. $\neq$ dimension of R.H.S.
Hence, the given equation is dimensionally incorrect.