# integration (squareroot)

#### gomes

What would be the best/quickest way to integrate

squareroot(1-x^2) ?

#### earboth

MHF Hall of Honor
What would be the best/quickest way to integrate

squareroot(1-x^2) ?
1. Use integration by parts:

$$\displaystyle \int(\sqrt{1-x^2}) dx = \int(1 \cdot \sqrt{1-x^2}) dx$$

2. In the second step keep in mind that

$$\displaystyle \int \left(\frac1{\sqrt{1-x^2}}\right) dx = \arcsin(x)$$

#### tonio

What would be the best/quickest way to integrate

squareroot(1-x^2) ?

Substitute $$\displaystyle \sin t=x$$ ...(Wink)

Tonio

• HallsofIvy

#### tonio

1. Use integration by parts:

$$\displaystyle \int(\sqrt{1-x^2}) dx = \int(1 \cdot \sqrt{1-x^2}) dx$$

2. In the second step keep in mind that

$$\displaystyle \int \left(\frac1{\sqrt{1-x^2}}\right) dx = \arcsin(x)$$

I don't think the above helps: $$\displaystyle u=\sqrt{1-x^2}\Longrightarrow u'=-\frac{x}{\sqrt{1-x^2}}\,,\,\,v'=1\Longrightarrow v=x$$ , so we get the integral $$\displaystyle \int\frac{x^2}{\sqrt{1-x^2}}\,dx$$ ...which not only is not an arcsine but it doesn't look easy at all...

Unless, of course, I missed something.

Tonio

#### Prove It

MHF Helper
What would be the best/quickest way to integrate

squareroot(1-x^2) ?
Trigonometric or Hyperbolic substitution works well...

Let $$\displaystyle x = \sin{\theta}$$ so that $$\displaystyle dx = \cos{\theta}\,d\theta$$.

The integral becomes

$$\displaystyle \int{\sqrt{1 - x^2}\,dx} = \int{\sqrt{1 - \sin^2{\theta}}\,\cos{\theta}\,d\theta}$$

$$\displaystyle = \int{\cos{\theta}\cos{\theta}\,d\theta}$$

$$\displaystyle = \int{\cos^2{\theta}\,d\theta}$$

$$\displaystyle = \int{\frac{1}{2}\cos{2\theta} + \frac{1}{2}\,d\theta}$$

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

$$\displaystyle = \frac{2\sin{\theta}\cos{\theta}}{4} + \frac{\theta}{2} + C$$

$$\displaystyle = \frac{\sin{\theta}\cos{\theta} + \theta}{2} + C$$.

Remembering that $$\displaystyle x = \sin{\theta}$$, that means $$\displaystyle \cos{\theta} = \sqrt{1 - x^2}$$ and $$\displaystyle x = \arcsin{\theta}$$ and substituting, we find

$$\displaystyle \int{\sqrt{1 - x^2}\,dx} = \frac{x\sqrt{1 - x^2} + \arcsin{x}}{2} + C$$.

• HallsofIvy

#### [email protected]

Unless, of course, I missed something.

Tonio
I think earboth is going to re-write

$$\displaystyle \frac{-x^2}{\sqrt{1 - x^2}}$$

as

$$\displaystyle \frac{1 - x^2 - 1}{\sqrt{1 - x^2}}\ =\ \sqrt{1 - x^2}\ -\ \frac{1}{\sqrt{1 - x^2}}$$

and then solve the top row of... ... for I. Which is no longer, really.

Key: ... is lazy integration by parts, doing without u and v. ... is the product rule, straight continuous lines differentiating downwards (integrating up) with respect to x.

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• earboth

#### gomes

Thanks everyone, really appreciate it. 