Question

$\int \frac{dx}{\sqrt{2ax-x^2}}$ = $a^{n}si{n}^{-1}[\frac{x}{a}-1]$

The value of n is

• A. 0
• B. -1
• C. 1
• D. None of these

Answer 1

The correct option is A

0

We can see that 2ax and $x^2$ need to have the same dimension as they are being added.

$\therefore \left[2ax\right]=L^2$

$\Rightarrow \left[a\right]=L$

$LHS=\int \frac{dx}{\sqrt{2ax-x^2}}$

$\left[LHS\right]=\frac{L}{\sqrt{L^2-L^2}}=L^0$

$si{n}^{-1}[\frac{x}{a}-1]$ is dimensionless as it gives an angle

$\therefore \left[RHS\right]=\left[a^n\right]=L^n$

$\therefore \left[LHS\right]=\left[RHS\right]$

$\Rightarrow L^0=L^n$

$\Rightarrow n=0$