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

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

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Solution

(a) 0

[ax] = [x²]
⇒ [a] = [x] ...(1)

Dimension of LHS = Dimension of RHS
$\Rightarrow [\frac{d x}{\sqrt{x^{2}}}] = [a^{n}] \Rightarrow [\frac{L}{L}] = [a^{n}] . . . (2) [L^{0}] = [a^{n}] n = 0$

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Similar questions

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

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.

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

The value of n is:

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

The value of n is

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

n

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