Question

integral dx/(2ax-x^2)^1/2=a^n sin^-1[x/a-1] ∫dx/√2ax-x² =aⁿ sin^-1 [x/a-1] the value of 'n' by dimensional analyses is

Answer 1

Dear student,

Let dimension of x, $\left[x\right]=\mathrm{L}$
So, by dimension analysis,
$$\begin{gathered} \left[ax\right]=\left[x^2\right]={\mathrm{L}}^2 \\ \Rightarrow \left[a\right]=\mathrm{L} \\ I\mathrm{ntegration}\mathrm{sign}\mathrm{do}\mathrm{not}\mathrm{affect}\mathrm{dimensions}. \\ \mathrm{So},\mathrm{dimension}\mathrm{of}\mathrm{LHS}, \\ \frac{\left[x\right]}{\sqrt{\left[x^2\right]}}=\frac{\mathrm{L}}{\mathrm{L}}=dimensionless, \\ \mathrm{Thus},\mathrm{RHS}\mathrm{is}\mathrm{also}\mathrm{dimensionless}, \\ \mathrm{So},\mathrm{n}=0 \end{gathered}$$

Regards.