integration of [dx/(2ax-x²)^0.5] = aⁿ sin[x/a-1]; where a and x are distance ,find the value of n using dimension al analysis

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Solution

$\frac{d x}{\sqrt{2 a x - x^{2}}} = \frac{1}{\sqrt{[L^{2}]}} = [L]^{- 1} - - - - - - - - (i) a^{n} \sin (\frac{x}{a} - 1) = [L]^{n} - - - - - - - - (i i) c o m p a i r i n g i a n d i i e q u a t i o n w e g e t n = - 1$

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\int \frac{d x}{\sqrt{2 a x - x^{2}}} = a^{n} {s i n}^{- 1} [\frac{x - a}{a}]

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\int \frac{d x}{{(2 a x - x^{2})}^{1 / 2}} = a^{n} s i n^{- 1} (\frac{x}{a} - 1)

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Int dx/√2ax-x,^2 = a^n sin^-1[x/a-1] .

The value of n is
(a) 0 (b) —1
(c) 1 (d) none of these.
You may use dimensional analysis to solve the problem.

\int \frac{x d x}{\sqrt{2 a x - x^{2}}} = a^{n} {sin}^{- 1} [\frac{x}{a} - 1]

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