How do you integrate $\int \frac{6 - x^{2}}{\sqrt{9 - x^{2}}} d x$ $int (6-x^2)/sqrt(9-x^2) dx$ using trigonometric substitution?

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How do you integrate $\int \frac{6 - x^{2}}{\sqrt{9 - x^{2}}} d x$ $int (6-x^2)/sqrt(9-x^2) dx$ using trigonometric substitution?

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Answer 1 · 2016-11-08T17:31:38

$\frac{1}{2} x \sqrt{9 - x^{2}} + \frac{3}{2} arcsin (\frac{x}{3}) + C$ $1/2xsqrt(9-x^2)+3/2arcsin(x/3)+C$

Explanation:

$I = \int \frac{6 - x^{2}}{\sqrt{9 - x^{2}}} d x$ $I=int(6-x^2)/sqrt(9-x^2)dx$

Rewrite the integral for simplification:

$I = \int \frac{- 3 + 9 - x^{2}}{\sqrt{9 - x^{2}}} d x = - 3 \int \frac{d x}{\sqrt{9 - x^{2}}} + \int \frac{9 - x^{2}}{\sqrt{9 - x^{2}}} d x$ $I=int(-3+9-x^2)/sqrt(9-x^2)dx=-3intdx/sqrt(9-x^2)+int(9-x^2)/sqrt(9-x^2)dx$

$I = - 3 \int \frac{d x}{\sqrt{9 - x^{2}}} + \int \sqrt{9 - x^{2}} d x$ $color(white)I=-3intdx/sqrt(9-x^2)+intsqrt(9-x^2)dx$

First focusing on $J = \int \frac{d x}{\sqrt{9 - x^{2}}}$ $J=intdx/sqrt(9-x^2)$ . For this, let $x = 3 sin θ$ $x=3sintheta$ . This implies that $d x = 3 cos θ d θ$ $dx=3costhetad theta$ .

$J = \int \frac{d x}{\sqrt{9 - x^{2}}} = \int \frac{3 cos θ d θ}{\sqrt{9 - 9 {sin}^{2} θ}} = \int \frac{cos θ}{\sqrt{1 - {sin}^{2} θ}} d θ = \int d θ$ $J=intdx/sqrt(9-x^2)=int(3costhetad theta)/sqrt(9-9sin^2theta)=intcostheta/sqrt(1-sin^2theta)d theta=intd theta$

$J = θ = arcsin (\frac{x}{3})$ $color(white)J=theta=arcsin(x/3)$

Recall that the $θ = arcsin (\frac{x}{3})$ $theta=arcsin(x/3)$ relationship comes from our earlier substitution $x = 3 sin θ$ $x=3sintheta$ .

Now we can work on the next integral $K = \int \sqrt{9 - x^{2}} d x$ $K=intsqrt(9-x^2)dx$ . We will use the same substitution as before, $x = 3 sin ϕ$ $x=3sinphi$ , implying that $d x = 3 cos ϕ d ϕ$ $dx=3cosphidphi$ .

$K = \int \sqrt{9 - x^{2}} d x = \int \sqrt{9 - 9 {sin}^{2} ϕ} (3 cos ϕ d ϕ)$ $K=intsqrt(9-x^2)dx=intsqrt(9-9sin^2phi)(3cosphidphi)$

$K = 9 \int \sqrt{1 - {sin}^{2} ϕ} (cos ϕ) d ϕ = 9 \int {cos}^{2} ϕ d ϕ$ $color(white)K=9intsqrt(1-sin^2phi)(cosphi)dphi=9intcos^2phidphi$

Rewrite by solving for ${cos}^{2} x$ $cos^2x$ in the identity $cos 2 x = 2 {cos}^{2} x - 1$ $cos2x=2cos^2x-1$ .

$K = 9 \int \frac{cos 2 ϕ + 1}{2} d ϕ = \frac{9}{2} \int cos 2 ϕ d ϕ + \frac{9}{2} \int d ϕ$ $K=9int(cos2phi+1)/2dphi=9/2intcos2phidphi+9/2intdphi$

For the first integral, we can use the substitution $u = 2 ϕ$ $u=2phi$ so $d u = 2 d ϕ$ $du=2dphi$ and $d ϕ = \frac{1}{2} d u$ $dphi=1/2du$ . The second integral is a fundamental one:

$K = \frac{9}{4} \int cos u d u + \frac{9}{2} ϕ = \frac{9}{4} sin u + \frac{9}{2} ϕ = \frac{9}{4} sin 2 ϕ + \frac{9}{2} ϕ$ $K=9/4intcosudu+9/2phi=9/4sinu+9/2phi=9/4sin2phi+9/2phi$

Using $sin 2 x = 2 sin x cos x$ $sin2x=2sinxcosx$ :

$K = \frac{9}{2} sin ϕ cos ϕ + \frac{9}{2} ϕ$ $K=9/2sinphicosphi+9/2phi$

Now to undo these substitutions, recall that $x = 3 sin ϕ$ $x=3sinphi$ . Thus, $ϕ = arcsin (\frac{x}{3})$ $phi=arcsin(x/3)$ and $sin ϕ = \frac{x}{3}$ $sinphi=x/3$ . Thus $cos ϕ = \sqrt{1 - {sin}^{2} ϕ} = \sqrt{1 - \frac{x^{2}}{9}} = \frac{1}{3} \sqrt{9 - x^{2}}$ $cosphi=sqrt(1-sin^2phi)=sqrt(1-x^2/9)=1/3sqrt(9-x^2)$ . So:

$K = \frac{9}{2} (\frac{x}{3}) (\frac{1}{3} \sqrt{9 - x^{2}}) + \frac{9}{2} arcsin (\frac{x}{3})$ $K=9/2(x/3)(1/3sqrt(9-x^2))+9/2arcsin(x/3)$

$K = \frac{1}{2} x \sqrt{9 - x^{2}} + \frac{9}{2} arcsin (\frac{x}{3})$ $color(white)K=1/2xsqrt(9-x^2)+9/2arcsin(x/3)$

Returning to the original integral:

$I = - 3 J + K$ $I=-3J+K$

$I = - 3 arcsin (\frac{x}{3}) + [\frac{1}{2} x \sqrt{9 - x^{2}} + \frac{9}{2} arcsin (\frac{x}{3})]$ $color(white)I=-3arcsin(x/3)+[1/2xsqrt(9-x^2)+9/2arcsin(x/3)]$

$I = \frac{1}{2} x \sqrt{9 - x^{2}} + \frac{3}{2} arcsin (\frac{x}{3}) + C$ $color(white)I=1/2xsqrt(9-x^2)+3/2arcsin(x/3)+C$

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How do you integrate $\int \frac{6 - x^{2}}{\sqrt{9 - x^{2}}} d x$ $int (6-x^2)/sqrt(9-x^2) dx$ using trigonometric substitution?

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