You may not know integration but using dimensional analysis, you can check on some results. In the original $\int{\dfrac{dx}{{{\left( 2ax-{{x}^{2}} \right)}^{\dfrac{1}{2}}}}}={{a}^{n}}{{\sin }^{-1}}\left( \dfrac{x}{a}-1 \right)$, find the values of n.

Hint: In order to figure out if the given integral is true or not, we can use dimensional analysis. If the LHS and RHS of the equation have the same dimension, then the equation can be taken as valid. Also, remember that we can only subtract or add terms of the same dimensions. Also, a number has no dimension, and the ratio of quantities having similar dimensions are dimensionless.

Complete step by step answer:
We need the dimensions for the LHS and the RHS in order to find the value of n. Suppose the term ‘x’ in the given equation denotes distance or length. The dimensional for length is $\left[ L \right]$. So, the integrand in the LHS of the equation is, $\dfrac{1}{{{\left( 2ax-{{x}^{2}} \right)}^{\dfrac{1}{2}}}}$.
In order to add $\text{2ax with }{{\text{x}}^{2}}$, 2ax should have the same dimension of ${{\text{x}}^{\text{2}}}$ , i.e. $\left[ {{L}^{2}} \right]$ in this case. In order to satisfy this condition, the term ‘a’ should have the dimension of length or the dimension of x. So, we arrived at the conclusion that ‘a’ has the same dimension of x. So, writing the LHS in the dimensional form we get,
$\dfrac{dx}{{{\left( 2ax-{{x}^{2}} \right)}^{\dfrac{1}{2}}}}=\dfrac{\left[ L \right]}{{{\left[ {{L}^{2}} \right]}^{\dfrac{1}{2}}}}$
$\therefore \dfrac{dx}{{{\left( 2ax-{{x}^{2}} \right)}^{\dfrac{1}{2}}}}=\left[ {{L}^{0}} \right]$
So, the LHS of the equation is dimensionless.
If the RHS is considered, we know that ${{\sin }^{-1}}\left( \dfrac{x}{a}-1 \right)$ is an angle, and an angle has no dimensions. Next one is the term ‘a’, according to our previous discussion ‘a’ should have the dimension of length, so we can write the dimensional formula as,
${{a}^{n}}={{\left[ L \right]}^{n}}$
But, the LHS of the equation is dimensionless, so, in order for the equation being valid, the RHS should also be dimensionless. In that case, the value of ‘n’ should be zero.
So, the value of n is zero.

Note: All valid equations will have the same dimension in its LHS and RHS. But it is not true that all dimensionally balanced equations are valid. This is because there may be terms missing from the LHS or RHS, which makes the equation valid.
The integral given in the question is easily solvable, and the answer will be ${{\sin }^{-1}}\left( \dfrac{x-a}{a} \right)$.
Physical quantities will always have units. A physical quantity can be dimensionless, but still, it will have a unit. An angle is an example of a dimensionless quantity that has a unit called radian.