int_{}^{} {frac{{dx}}{{{{(2ax - {x^2})}^{1/2} | vedclass

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Question

$x$ = length
$	herefore \,[X]\, = \,[L]$ and $ [dx]\, = \,[L]$
$\left[ {\frac{x}{a}} ight]\, =$ dimensionless 
$	herefore \,[a] = [x]\, =

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$0$

Vedclass · Answer

$x$ = length
$	herefore \,[X]\, = \,[L]$ and $ [dx]\, = \,[L]$
$\left[ {\frac{x}{a}} ight]\, =$ dimensionless 
$	herefore \,[a] = [x]\, = \,[L]$ $\frac{{[L]}}{{{{[{L^2} - {L^2}]}^{1/2}}}} = [{L^n}]$
$	herefore \,\,n\, = \,0$

$\int_{}^{} {\frac{{dx}}{{{{(2ax - {x^2})}^{1/2}}}} = {a^n}{{\sin }^{ - 1}}\left( {\frac{x}{a} - 1} \right)} $ in this formula $n =$ _____

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