# How do you find the integral of  sqrt (2x - x^2) dx?

Jul 4, 2018

The answer is $= \frac{1}{2} \arcsin \left(x - 1\right) + \frac{1}{2} \left(x - 1\right) \sqrt{2 x - {x}^{2}} + C$

#### Explanation:

$2 x - {x}^{2} = 1 - {\left(x - 1\right)}^{2}$ by completing the square.

Therefore, the integral is

$I = \int \sqrt{2 x - {x}^{2}} \mathrm{dx} = \int \sqrt{1 - {\left(x - 1\right)}^{2}} \mathrm{dx}$

Let $u = x - 1$, $\implies$, $\mathrm{du} = \mathrm{dx}$

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

Let $u = \sin \theta$, $\implies$, $\mathrm{du} = \cos \theta d \theta$

$\sqrt{1 - {u}^{2}} = \sqrt{1 - {\sin}^{2} \theta} = \cos \theta$

$I = \int \cos \theta \cdot \cos \theta d \theta = \int {\cos}^{2} \theta d \theta$

$\cos 2 \theta = 2 {\cos}^{2} \theta - 1$

$\implies$, ${\cos}^{2} \theta = \frac{1 + \cos 2 \theta}{2}$

Therefore,

$I = \frac{1}{2} \int \left(1 + \cos 2 \theta\right) d \theta$

$= \frac{1}{2} \left(\theta + \frac{1}{2} \sin 2 \theta\right)$

$= \frac{1}{2} \theta + \frac{1}{4} \sin 2 \theta$

$= \frac{1}{2} \theta + \frac{1}{2} \sin \theta \cos \theta$

$= \frac{1}{2} \arcsin \left(u\right) + \frac{1}{2} u \sqrt{1 - {u}^{2}}$

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

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

Jul 4, 2018

$\arcsin \sqrt{\frac{x}{2}} - \frac{1}{2} \left(1 - x\right) \sqrt{2 x - {x}^{2}} + C$.

#### Explanation:

Here is Second Method :

Let, $x = 2 {\sin}^{2} u . \therefore \mathrm{dx} = 2 \cdot 2 \sin u \cdot \cos u \mathrm{du}$.

$\therefore \int \sqrt{2 x - {x}^{2}} \mathrm{dx}$,

$= \int \sqrt{4 {\sin}^{2} u - 4 {\sin}^{4} u} \left(4 \sin u \cos u\right) \mathrm{du}$,

$= 4 \int \sqrt{4 {\sin}^{2} u \left(1 - {\sin}^{2} u\right)} \sin u \cos u \mathrm{du}$,

$= 4 \int \left(2 \sin u \cos u\right) \sin u \cos u \mathrm{du}$,

$= 2 \int \left(4 {\sin}^{2} u {\cos}^{2} u\right) \mathrm{du}$,

$= 2 \int {\sin}^{2} 2 u \mathrm{du}$,

$= 2 \int \frac{1 - \cos 4 u}{2} \mathrm{du}$,

$= u - \frac{1}{4} \sin 4 u$,

$= u - \frac{1}{4} \cdot 2 \sin 2 u \cos 2 u$,

$= u - \frac{1}{2} \left(2 \sin u \cos u\right) \left(1 - 2 {\sin}^{2} u\right)$,

$= u - \sqrt{{\sin}^{2} u {\cos}^{2} u} \left(1 - 2 {\sin}^{2} u\right)$,

$= u - \sqrt{{\sin}^{2} u \left(1 - {\sin}^{2} u\right)} \left(1 - 2 {\sin}^{2} u\right)$.

Since, $x = 2 {\sin}^{2} u , \sin u = \sqrt{\frac{x}{2}} . \therefore u = \arcsin \sqrt{\frac{x}{2}}$.

$\Rightarrow \int \sqrt{2 x - {x}^{2}} \mathrm{dx}$,

$= \arcsin \sqrt{\frac{x}{2}} - \sqrt{\frac{x}{2} \left(1 - \frac{x}{2}\right)} \left(1 - x\right)$,

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

Enjoy Maths.!