Integrate sin^-1 (√(x/(x+a))) dx

∫sin^-1[√{x/(x+a)} ] dx

integrate by parts

u = sin^-1[√{x/(x+a)} ]
du = 1/[√{1 - x/(x+a) ]* 1/2[√{x + a)/x]*(x + a - x} /(x + a)^2
= 1/[√{a/(x+a) ]* 1/2[√{x + a)/x]*( a) /(x + a)^2
= 1/2√{(x+a)/a}√{x + a)/x]*( a) /(x + a)^2
= (1/2)√(a/x)(1 /(x + a)) dx

dv = dx
v = x

∫sin^-1[√{x/(x+a)} ] dx = xsin^-1[√{x/(x+a)} ] - 1/2 ∫ x√(a/x)(1 /(x + a)) dx

= xsin^-1[√{x/(x+a)} ] - √a /2 ∫ √x dx /(x + a)

let √x = √a tan(u)

(1/2)(1/√x) dx = √a sec^2(u) du

dx = 2√(ax) sec^2(u) du

dx = 2a tan(u) sec^2(u) du

∫sin^-1[√{x/(x+a)} ] dx = xsin^-1[√{x/(x+a)} ] - √a /2 ∫√a tan(u)(2a tan(u) sec^2(u)) du /a sec^2(u)

= xsin^-1[√{x/(x+a)} ] - a ∫ tan^2(u) du

= xsin^-1[√{x/(x+a)} ] - a ∫ [sec^2(u) - 1 ] du

= xsin^-1[√{x/(x+a)} ] - a [tan(u) - u] + c

= xsin^-1[√{x/(x+a)} ] - a [√(x/a) - tan^-1(√(x/a) ] + c

= xsin^-1[√{x/(x+a)} ] - √(ax) - a tan^-1(√(x/a) ] + C

http://tinypic.com/r/313j1vs/7