`sqrt((a-x)/(x))`

Explanation

`Let I=int sqrt((a-x)/x)dx`
Put `x=a sin^(2) theta" "therefore dx=axx2 sin theta cos theta d theta and sin^(2)theta=(x)/(a)`
`therefore I=intsqrt((a-asin^(2)theta)/(a sin^(2)theta)).2a sin thetacos theta d theta` ltbvrgt `int sqrt((a(1-sin^(2)theta))/(a sin^(2)theta)).2a sin theta cos theta d theta=int (cos theta)/(sin theta) .2a sin theta cos theta d theta`
`=int 2a cos^(2)theta d theta=a int 2 cos^(2)theta d theta=a int(1+cos 2 theta)d theta `
`=a int 1 d theta+a int cos 2 theta d theta `
`=a theta+a.(sin 2 theta)/(2)+c`
`=a theta+a.(sin 2 theta cos theta)/(2)+c`
`=a theta+a sin theta cos theta+c .....(1)`
`Now sin^(2)theta=(x)/(a)" "therefore sin theta=sqrt((x)/(a)`
`therefore theta=sin^(-1)sqrt((x)/(a))`
`and cos theta=sqrt(1-sin^(2)theta)=sqrt(1-(x)/(a))=sqrt((a-x)/(a))`
`therefore` from (1)
`i=a sin^(-1) sqrt((x)/(a))+a .sqrt((x)/(a)).sqrt((a-x)/(a))+c`
`=a sin^(-1)sqrt((x)/(a))+sqrt(x(a-x))+c`