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How do you integrate $\int \frac{x^{2} - x}{\sqrt{9 - x^{2}}} d x$ $int (x^2 -x) / sqrt(9 - x^2) dx$ using trigonometric substitution?

1 Answer

May 13, 2018

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

Explanation:

We know that,

$(1) {sin}^{2} x = \frac{1 - cos 2 x}{2}$ $color(red)((1)sin^2x=(1-cos2x)/2)$

$(2) \int cos x d x = sin x + c$ $color(blue)((2)intcosxdx=sinx+c)$

$(3) \int sin x d x = - cos x + c$ $color(violet)((3)intsinxdx=-cosx+c)$

Here,

$I = \int \frac{x^{2} - x}{\sqrt{9 - x^{2}}} d x$ $I=int(x^2-x)/sqrt(9-x^2)dx$

Subst. $x = 3 sin u \Rightarrow d x = 3 cos u d u$ $x=3sinu=>dx=3cosudu$

$\Rightarrow x^{2} = 9 {sin}^{2} u and sin u = \frac{x}{3} \Rightarrow u = {sin}^{- 1} (\frac{x}{3})$ $=>x^2=9sin^2u and sinu=x/3=>u=sin^-1(x/3)$

So,

$I = \int \frac{9 {sin}^{2} u - 3 sin u}{\sqrt{9 - 9 {sin}^{2} u}} \times 3 cos u d u$ $I=int(9sin^2u-3sinu)/sqrt(9-9sin^2u)xx3cosudu$

$=int(9sin^2u-3sinu)/(cancel(3cosu))xxcancel(3cosu)du$

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

$=9int(1-cos(2u))/2du-3intsinudu...toApplycolor(blue)((2)) andcolor(violet)((3))$

$=9/2[u-sin(2u)/2]-3(-cosu)+c$

$=9/4[2u-sin2u]+3cosu+c$

Subst. back , $sinu=x/3 and u=sin^-1(x/3)$

$I=9/4[2sin^-1(x/3)-2sinucosu]+3sqrt(1-sin^2u)+c$

$=9/4xx2[sin^-1 (x/3)-sinusqrt(1-sin^2u)]+3sqrt(1-x^2/9)+c$

$=9/2[sin^-1(x/3)-(x/3)sqrt(1-x^2/9)]+3sqrt(9-x^2)/3+c$

$=9/2sin^-1(x/3)-9/2xxx/3sqrt(9-x^2)/3+sqrt(9-x^2)+c$

$I=9/2sin^-1(x/3)-x/2sqrt(9-x^2)+sqrt(9-x^2)+c$

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