# How do you integrate int (x^2 -x) / sqrt(9 - x^2) dx using trigonometric substitution?

May 13, 2018

$I = \frac{9}{2} {\sin}^{-} 1 \left(\frac{x}{3}\right) - \frac{x}{2} \sqrt{9 - {x}^{2}} + \sqrt{9 - {x}^{2}} + c$

#### Explanation:

We know that,

$\textcolor{red}{\left(1\right) {\sin}^{2} x = \frac{1 - \cos 2 x}{2}}$

$\textcolor{b l u e}{\left(2\right) \int \cos x \mathrm{dx} = \sin x + c}$

$\textcolor{v i o \le t}{\left(3\right) \int \sin x \mathrm{dx} = - \cos x + c}$

Here,

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

Subst. $x = 3 \sin u \implies \mathrm{dx} = 3 \cos u \mathrm{du}$

$\implies {x}^{2} = 9 {\sin}^{2} u \mathmr{and} \sin u = \frac{x}{3} \implies u = {\sin}^{-} 1 \left(\frac{x}{3}\right)$

So,

$I = \int \frac{9 {\sin}^{2} u - 3 \sin u}{\sqrt{9 - 9 {\sin}^{2} u}} \times 3 \cos u \mathrm{du}$

$= \int \frac{9 {\sin}^{2} u - 3 \sin u}{\cancel{3 \cos u}} \times \cancel{3 \cos u} \mathrm{du}$

=9intsin^2udu-3intsinudu...tocolor(red)(Apply(1)

$= 9 \int \frac{1 - \cos \left(2 u\right)}{2} \mathrm{du} - 3 \int \sin u \mathrm{du} \ldots \to A p p l y \textcolor{b l u e}{\left(2\right)} \mathmr{and} \textcolor{v i o \le t}{\left(3\right)}$

$= \frac{9}{2} \left[u - \sin \frac{2 u}{2}\right] - 3 \left(- \cos u\right) + c$

$= \frac{9}{4} \left[2 u - \sin 2 u\right] + 3 \cos u + c$

Subst. back , $\sin u = \frac{x}{3} \mathmr{and} u = {\sin}^{-} 1 \left(\frac{x}{3}\right)$

$I = \frac{9}{4} \left[2 {\sin}^{-} 1 \left(\frac{x}{3}\right) - 2 \sin u \cos u\right] + 3 \sqrt{1 - {\sin}^{2} u} + c$

$= \frac{9}{4} \times 2 \left[{\sin}^{-} 1 \left(\frac{x}{3}\right) - \sin u \sqrt{1 - {\sin}^{2} u}\right] + 3 \sqrt{1 - {x}^{2} / 9} + c$

$= \frac{9}{2} \left[{\sin}^{-} 1 \left(\frac{x}{3}\right) - \left(\frac{x}{3}\right) \sqrt{1 - {x}^{2} / 9}\right] + 3 \frac{\sqrt{9 - {x}^{2}}}{3} + c$

$= \frac{9}{2} {\sin}^{-} 1 \left(\frac{x}{3}\right) - \frac{9}{2} \times \frac{x}{3} \frac{\sqrt{9 - {x}^{2}}}{3} + \sqrt{9 - {x}^{2}} + c$

$I = \frac{9}{2} {\sin}^{-} 1 \left(\frac{x}{3}\right) - \frac{x}{2} \sqrt{9 - {x}^{2}} + \sqrt{9 - {x}^{2}} + c$