[tex] \int\limits^{}_{} {\sqrt{2 a x - x^2}} \, dx =a^n*sin^{-1}(\frac{x}{a}-1)\\By\ dimensional\ analysis,LHS\ has\ a\ dim\ of\ x^2\ or\ a^2.\\\\Hence, n=2.\ \ Sin^{-1}\ does\ have\ any\ dim.[/tex]
[tex]I= \int\limits^{}_{} {\sqrt{2 a x - x^2}} \, dx =a^n*sin^{-1}(\frac{x}{a}-1)\\\\Let,\ \ x=2aCos \theta\\\\I=\int\limits^{}_{} {\sqrt{2 a x - x^2}} \, dx = \int\limits^{}_{} {\sqrt{4 a^2 Cos \theta - 4a^2Cos^2\theta}} *(-2aSin\theta)\, d\theta\\\\=-4a^2\int\limits^{}_{} {\sqrt{Cos \theta -Cos^2\theta}} *Sin\theta\, d\theta[/tex]
[tex]I=-4a^2\int\limits^{}_{} {(\sqrt{Cos \theta} \sqrt{1 -Cos\theta}} *Sin\theta\, d\theta\\\\=-4a^2\int\limits^{}_{} {(\sqrt{Cos \theta} \sqrt2 Sin(\theta/2)} *Sin\theta\, d\theta\\\\=-4a^2[/tex]
