#I=intarcCos sqrt(a/(a+x))dx=intcos^-1sqrt(a/(a+x))dx#
Let, #x=atan^2u=>dx=2atanusec^2udu#
#andcolor(red)
(tan^2u=x/a=>tanu=sqrt(x/a)=>u=tan^-1sqrt(x/a)to(1)#
So,
#I=intcos^-1sqrt(a/(a+atan^2u))xx2atanusec^2udu#
#=2aintcos^-1sqrt(a/(asec^2u))tanusec^2udu#
#=2aintcos^-1sqrt(cos^2u)tanusec^2udu#
#=2aintcos^-1(cosu)tanusec^2udu#
#I=2aint(u)(tanusec^2u)du#
#"Using "color(blue)"Integration by Parts :"#
We note that, #inttanusec^2udu=inttanud/(du)(tanu)du=tan^2u/2#
So,
#I=2a[uinttanusec^2udu-int(1xxinttanusec^2udu)du]#
#=2a[utan^2u/2-inttan^2u/2du]#
#=a[utan^2u-int(sec^2u-1)du]#
#I=a[utan^2u-tanu+u]+c#
Hence, from #color(red)((1)#,we get
#I=a[tan^-1sqrt(x/a)xx x/a-sqrt(x/a)+tan^-1sqrt(x/a)]+c#
#I=a[x/atan^-1sqrt(x/a)+tan^-1sqrt(x/a)-sqrt(x/a)]+c#
#I=a[(x/a+1)tan^-1sqrt(x/a)-sqrt(x/a)]+c#
#I=(x+a)tan^-1sqrt(x/a)-sqrt(ax)+c#
