Question

What is `int 1/(sqrt[x]*sqrt[4-x]) dx`?

Answer 1

`2sin^(-1)(sqrtx/2)+c`

Explanation:

`I=int 1/(sqrt[x]*sqrt[4-x]) dx`?

substitute `u=sqrtx/2=>x=4u^2`

`=>du=1/4x^(-1/2)`

`:. dx=4sqrtxdu`

`I=int1/(cancel(sqrtx)(sqrt(4-4u^2)))4cancel(sqrtx)du`

`I=int4/(2sqrt(1-u^2))du`

`=2int(du)/(sqrt(1-u^2))`

this is astandard integral

`=2sin^(-1)u+c`

`=2sin^(-1)(sqrtx/2)+c`

Answer 2

`=-2sin^-1(sqrt(4-x)/2)+C`

Explanation:

We know that,

`(I)color(red)(int1/sqrt(a^2-x^2)dx=sin^-1(x/a)+c`

Here,

`I=int1/(sqrtx*sqrt(4-x))dx`

Let, `sqrt(4-x)=u=>4-x=u^2`

i.e. `x=4-u^2=>dx=-2udu`

So,

`I=int(-2u)/(sqrt(4-u^2)*u)du`

`=-2int1/sqrt(2^2-u^2)du...toApply (I)`

`=-2sin^-1(u/2)+C,where,u=sqrt(4-x)`

`=-2sin^-1(sqrt(4-x)/2)+C`