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

Apr 6, 2018

$2 {\sin}^{- 1} \left(\frac{\sqrt{x}}{2}\right) + c$

#### Explanation:

$I = \int \frac{1}{\sqrt{x} \cdot \sqrt{4 - x}} \mathrm{dx}$?

substitute $u = \frac{\sqrt{x}}{2} \implies x = 4 {u}^{2}$

$\implies \mathrm{du} = \frac{1}{4} {x}^{- \frac{1}{2}}$

$\therefore \mathrm{dx} = 4 \sqrt{x} \mathrm{du}$

$I = \int \frac{1}{\cancel{\sqrt{x}} \left(\sqrt{4 - 4 {u}^{2}}\right)} 4 \cancel{\sqrt{x}} \mathrm{du}$

$I = \int \frac{4}{2 \sqrt{1 - {u}^{2}}} \mathrm{du}$

$= 2 \int \frac{\mathrm{du}}{\sqrt{1 - {u}^{2}}}$

this is astandard integral

$= 2 {\sin}^{- 1} u + c$

$= 2 {\sin}^{- 1} \left(\frac{\sqrt{x}}{2}\right) + c$

Apr 6, 2018

$= - 2 {\sin}^{-} 1 \left(\frac{\sqrt{4 - x}}{2}\right) + C$

#### Explanation:

We know that,

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

Here,

$I = \int \frac{1}{\sqrt{x} \cdot \sqrt{4 - x}} \mathrm{dx}$

Let, $\sqrt{4 - x} = u \implies 4 - x = {u}^{2}$

i.e. $x = 4 - {u}^{2} \implies \mathrm{dx} = - 2 u \mathrm{du}$

So,

$I = \int \frac{- 2 u}{\sqrt{4 - {u}^{2}} \cdot u} \mathrm{du}$

$= - 2 \int \frac{1}{\sqrt{{2}^{2} - {u}^{2}}} \mathrm{du} \ldots \to A p p l y \left(I\right)$

$= - 2 {\sin}^{-} 1 \left(\frac{u}{2}\right) + C , w h e r e , u = \sqrt{4 - x}$

$= - 2 {\sin}^{-} 1 \left(\frac{\sqrt{4 - x}}{2}\right) + C$