You may not know integration, but using dimensional analysis you can check on some results. In the original $\int \frac{d x}{{(2 a x - x^{2})}^{1 / 2}} = a^{n} s i n^{- 1} (\frac{x}{a} - 1)$ , find the value of $n$ .

Open in App

Solution

terms in trigonometric functions must be dimensionless, so x, a have same dimensions. on left hand side(LHS) if we take root of the dimension we will have same units as that of dx so now LHS is dimensionless RHS must be dimensionless. so power 'n' of 'a' must be zero.

Suggest Corrections

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} \sin^{- 1} [\frac{x}{a} - 1]

\int \frac{d x}{\sqrt{2 a x - x^{2}}} = a^{n} {sin}^{- 1} (\frac{x}{a} - 1) .

Join BYJU'S Learning Program