Question

Integration dx/√2ax-x^2=a^n sin^-1[x/a-1] the value of n by only using dimensional analysis

Answer 1

$\huge{\bold{\underline{\underline{....Answer....}}}}$

$\huge{\bold{\underline{Given:-}}}$

• "x" has a dimension of length.

$\huge{\bold{\underline{Explanation:-}}}$

$\large{\displaystyle\int \dfrac{dx}{ \sqrt{2ax - x^2}} = a^n \sin^{-1} \left[\dfrac{x}{a} - 1 \right]}$

By the principle of homogeneity,

$\large{ax = x^2}$

Note:-

• Here "a" is not acceleration
• negative sign should be neglected.
• And, 2 can also be neglected as it is dimension less.

Now,

$\large{[ax] = [x]^2}$

$\large{[a] = \dfrac{x^2}{x}}$

$\large{[a] = \dfrac{ \cancel{x ^{2}}}{ \cancel{x}}}$

$\large{ \implies[a] = x}$

As x has a dimension of length,

$\large{ \implies [a] = L}$

Now, Dimensions at RHS = Dimensions at LHS

$\large{ \dfrac{dx}{ \sqrt{x^2}} = [a]^n}$

Here on doing integration the dimensions will not change.

Now,

• "dx" has a dimension of length .
• a has a dimension of length (Proved above).
• Here sine angle is neglected as it is dimension less.

$\large{ \dfrac{[L]}{ [\sqrt{L^2}]} = [L]^n}$

Now,

$\large{ \dfrac{[L]}{[L]} = [L]^n}$

$\large{[L]^{1-1} = [L]^n}$

$\large{[L]^0 = [L]^n}$

As both the Quantities are same

we compare it,

we get,

$\huge{\boxed{\boxed{n = 0}}}$

So, the value of n is Zero.

Note:-

All the calculations are done by dimensional analysis.

As specified by the question.