dx/√2ax-x^2=a^nsin^-1[x/a-1].The value of n is?you may use dimensional analysis to solve the problem.

And:- the value of n is 0 but how ?

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Solution

All the trigonometric function is dimensionless. Since both sides of sin^-1 function is dimension less. It means M^0L^0T^0 = 1
So, 1 = a^n
a^0 = a^n
n = 1

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Similar questions

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]

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

n

\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} {s i n}^{- 1} [\frac{x - a}{a}]

a =

n

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