Answer 1

Let `"I"=int_0^asqrt(x^3/(a^3-x^3))dx`

Put `x^3=a^3t  =>x=at^(1/3)`
Diff. w.r.t. x ,

`dx=a/3t^((-2)/3) dt`

Limits becomes →[𝟎,𝟏]

`"I"=int_0^1(t)^(3/2)(1-t)^(3/2)t^((-2)/3)a/3dt`

`=a/3int_0^1t^(5/6)(1-t)^(3/2)dt`

`=a/3beta(5/6,3/2)`

`"I"=a(sqrtxgamma(5/6))/(gamma(1/3)` …………{ from the definition of beta function}