# How do you find the indefinite integral of int x/(sqrt(9-x^2))?

Dec 3, 2016

$x {\sin}^{- 1} \left(\frac{x}{3}\right) - {\sin}^{- 1} \left(\frac{x}{3}\right) + c$.
By parts. Set $u \left(x\right) = x$, $\frac{\mathrm{dv}}{\mathrm{dx}} = \frac{1}{\sqrt{9 - {x}^{2}}}$

#### Explanation:

Then $\frac{\mathrm{du}}{\mathrm{dx}} = 1$ and $v = {\sin}^{- 1} \left(\frac{x}{3}\right)$.
The integration by parts formula is:
$\int u \left(x\right) v \left(x\right) \mathrm{dx} = u \left(x\right) v \left(x\right) - \int v \left(x\right) \frac{\mathrm{du}}{\mathrm{dx}} \mathrm{dx}$

(or $\int u \mathrm{dv} = u v - \int v \mathrm{du}$ if your prefer)

so that $v \left(x\right) = {\sin}^{- 1} \left(\frac{x}{3}\right)$
giving the integral shown in the answer.

Notes:

1. I assume that either you can quote the standard integral $\int \frac{1}{\sqrt{{a}^{2} - {x}^{2}}} \mathrm{dx} = {\sin}^{- 1} \left(\frac{x}{a}\right)$ (or ${\cos}^{- 1} \left(\frac{x}{a}\right)$, it makes little difference). If not, you have to remember that you can differentiate all the inverse trig functions e.g. $y = {\sin}^{- 1} x$ by getting the $x$ on to the left first: $x = \sin y$, then differentiate with respect to $y$ not $x$, getting $\frac{\mathrm{dx}}{\mathrm{dy}} = \cos y = \sqrt{1 - {x}^{2}}$, then reciprocate both sides dy/{dx}=1/(sqrt(1-x^2). Then the integration is reverse of differentiation.

2. You have to guess correctly which part to make $u \left(x\right)$ and which to make $v \left(x\right)$. A simple rule of thumb here is that that the $x$ on the top is an irritation and so if you differentiate it will turn into 1 and go away. If you guess wrong the $x$ will turn into an ${x}^{2}$ which is bad news. Similarly if the $x$ had been and ${x}^{2}$ you would have needed two rounds of integration by parts.

3. I gloss over certain issues about "principal value" and the choice of ${\sin}^{-} 1 x$ or ${\cos}^{-} 1 x$.

Dec 3, 2016

$\int \frac{x}{\sqrt{9 - {x}^{2}}} \mathrm{dx} = - \sqrt{9 - {x}^{2}} + C$

#### Explanation:

$I = \int \frac{x}{\sqrt{9 - {x}^{2}}} \mathrm{dx}$

Let $u = 9 - {x}^{2}$. Differentiating this shows that $\mathrm{du} = - 2 x \mathrm{dx}$. Luckily our numerator is this only off by a factor of $- 2$.

$I = - \frac{1}{2} \int \frac{- 2 x}{\sqrt{9 - {x}^{2}}} \mathrm{dx} = - \frac{1}{2} \int \frac{1}{\sqrt{u}} \mathrm{du} = - \frac{1}{2} \int {u}^{- \frac{1}{2}} \mathrm{du}$

Using $\int {u}^{n} \mathrm{du} = {u}^{n + 1} / \left(n + 1\right) + C$, this becomes:

$I = - \frac{1}{2} \left({u}^{\frac{1}{2}} / \left(\frac{1}{2}\right)\right) = - \sqrt{u} = - \sqrt{9 - {x}^{2}} + C$