Question

/ [( 2ax - x^2) ^1/2 ] dx = a^n ( sin^-1(x/a - 1 ))

/ stands for integral of

What is value of n

(A) 0
(B) -1
(C) 1
(D) none

You may use dimensional analysis

Answer 1

$\int\limits^{}_{} {\sqrt{2 a x - x^2}} \, dx =a^n*sin^{-1}(\frac{x}{a}-1)\\By\ dimensional\ analysis,LHS\ has\ a\ dim\ of\ x^2\ or\ a^2.\\\\Hence, n=2.\ \ Sin^{-1}\ does\ have\ any\ dim.$

$I= \int\limits^{}_{} {\sqrt{2 a x - x^2}} \, dx =a^n*sin^{-1}(\frac{x}{a}-1)\\\\Let,\ \ x=2aCos \theta\\\\I=\int\limits^{}_{} {\sqrt{2 a x - x^2}} \, dx = \int\limits^{}_{} {\sqrt{4 a^2 Cos \theta - 4a^2Cos^2\theta}} *(-2aSin\theta)\, d\theta\\\\=-4a^2\int\limits^{}_{} {\sqrt{Cos \theta -Cos^2\theta}} *Sin\theta\, d\theta$

$I=-4a^2\int\limits^{}_{} {(\sqrt{Cos \theta} \sqrt{1 -Cos\theta}} *Sin\theta\, d\theta\\\\=-4a^2\int\limits^{}_{} {(\sqrt{Cos \theta} \sqrt2 Sin(\theta/2)} *Sin\theta\, d\theta\\\\=-4a^2$