Question

int dx sqrt 2ax-x^ 2 =a^ n sin^ -1 [ x a -1] The value of n is (a) 0
(c) 1
(b)-1
(d) none of these.
You may use dimensional analysis to solve the problem.​

Answer 1

Given Integrand,

$\displaystyle \sf \int \dfrac{dx}{ \sqrt{2ax - {x}^{2} } } \\ \\ \longrightarrow \displaystyle \sf \int \dfrac{dx}{ \sqrt{ {a}^{2} - {a}^{2} + 2ax - {x}^{2} } } \\ \\ \longrightarrow \displaystyle \sf \int \dfrac{dx}{ \sqrt{ {a}^{2} - ( {a}^{2} - 2ax + {x}^{2} )} } \\ \\ \longrightarrow \displaystyle \sf \int \dfrac{dx}{ \sqrt{ {a}^{2} - (x - a) {}^{2} } }$

We know that,

$\star \displaystyle \boxed{ \boxed{ \sf \int \dfrac{dx}{ {a}^{2} - {x}^{2} } = {sin}^{ - 1} \big( \dfrac{x}{a} \big) }}$

Therefore,

$\longrightarrow \sf sin {}^{ - 1} \big( \dfrac{ x - a}{a} \big) \\ \\ \longrightarrow \sf\displaystyle \sf \int \dfrac{dx}{ \sqrt{2ax - {x}^{2} } } = sin {}^{ - 1} \big( \dfrac{ x}{a} - 1\big) + c - - - - - - - (1)$

According to the question,

$\displaystyle \sf \int \dfrac{dx}{ \sqrt{2ax - {x}^{2} } } = {a}^{n} \sf sin {}^{ - 1} \big( \dfrac{ x }{a} - 1 \big) - - - - - - - (2)$

Comparing equations (1) and (2), we can deduce :

$\sf {a}^{n} \sf sin {}^{ - 1} \big( \dfrac{ x }{a} - 1 \big) =sin {}^{ - 1} \big( \dfrac{ x}{a} - 1\big) \\ \\ \implies \sf {a}^{n} = 1 \\ \\ \implies \boxed{ \boxed{ \sf n = 0}}$

Option (A) is correct.