You may not know integration. But using dimensional analysis you can check on some results. In the integral ∫ dx/(2ax-x²)⁰.⁵ = aⁿ sin⁻¹((x/a)-1) the value of n is

The correct option is 0.

Let us analyze the dimensions of both sides of the given equation to find the value of $n$ using dimensional analysis.

The given equation is:
$��\frac{dx}{{(2ax-x^2)}^{0.5}}=a^n{sin}^{-1}(\frac{x}{a}-1)$

Here, $x$ represents a position or distance, which has the dimension of length, denoted as:
$\left[x\right]=L$

Since $dx$ is an infinitesimal change in $x$, its dimension is also:
$\left[dx\right]=L$

In the term $2ax-x^2$, we are subtracting $x^2$ from $2ax$. For this subtraction to be dimensionally valid, both terms must have the same dimensions. Since $\left[x^2\right]=L^2$, the term $2ax$ must also have the dimension of $L^2$:
$\left[2ax\right]=\left[a\right]\left[x\right]=\left[a\right]L=L^2$
This gives the dimension of the constant $a$ as:
$\left[a\right]=L$

Now, let us determine the dimensions of the Left-Hand Side (LHS) of the equation:
$\left[\text{LHS}\right]=[\int \frac{dx}{{(2ax-x^2)}^{0.5}}]$
The integration symbol acts as a summation and does not change the overall dimensions. Thus:
$\left[\text{LHS}\right]=\frac{\left[dx\right]}{{[2ax-x^2]}^{0.5}}=\frac{L}{{\left(L^2\right)}^{0.5}}=\frac{L}{L}=L^0$
So, the LHS is a dimensionless quantity (dimension $L^0$).

Next, let us analyze the Right-Hand Side (RHS) of the equation:
$\left[\text{RHS}\right]=\left[a^n{sin}^{-1}(\frac{x}{a}-1)\right]$
Trigonometric functions and inverse trigonometric functions (like ${sin}^{-1}\theta$) represent angles and are dimensionless. Therefore:
$\left[{sin}^{-1}(\frac{x}{a}-1)\right]=L^0$
Consequently, the dimension of the RHS is determined solely by $a^n$:
$\left[\text{RHS}\right]={\left[a\right]}^n=L^n$

For the equation to be dimensionally correct, the dimensions of the LHS and RHS must be equal:
$\left[\text{LHS}\right]=\left[\text{RHS}\right]$
$L^0=L^n$

Comparing the exponents of $L$ on both sides, we get:
$n=0$