How do you find the integral of $\frac{2 x - 3}{{(9 - x^{2})}^{0.5}} d x$ $(2x-3)/((9-x^2)^0.5)dx$ ?

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How do you find the integral of $\frac{2 x - 3}{{(9 - x^{2})}^{0.5}} d x$ $(2x-3)/((9-x^2)^0.5)dx$ ?

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Answer 1 · 2015-08-05T02:32:29

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

Explanation:

Start off by rewriting as follows

$\int \frac{2 x}{{(9 - x^{2})}^{\frac{1}{2}}} d x - \int \frac{3}{{(9 - x^{2})}^{\frac{1}{2}}} d x$ $int (2x)/(9-x^2)^(1/2)dx-int 3/(9-x^2)^(1/2)dx$

For the first integral we use a u substitution

Let $u = 9 - x^{2}$ $u=9-x^2$ then $\frac{d u}{d x} = - 2 x$ $(du)/dx=-2x$

$d x = \frac{d u}{- 2 x}$ $dx=(du)/(-2x)$

Make the substitution

$- \int \frac{2 x}{u^{\frac{1}{2}}} \frac{d u}{2 x} = - \int u^{- \frac{1}{2}} d u$ $-int (2x)/u^(1/2)(du)/(2x)=-int u^(-1/2)du$

Integrating we have

$- 2 u^{\frac{1}{2}}$ $-2u^(1/2)$

Back substituting for u

$- 2 {(9 - x^{2})}^{\frac{1}{2}}$ $-2(9-x^2)^(1/2)$

$- 2 \sqrt{9 - x^{2}}$ $-2sqrt(9-x^2)$

For the second integral we will use a trigonometric substitution

Let $x = 3 sin θ$ $x=3sintheta$ the

$d x = 3 cos θ d θ$ $dx=3costhetad theta$

Make the substitution

$\int \frac{3}{{(9 - 9 {sin}^{2} θ)}^{\frac{1}{2}}} 3 cos d θ$ $int 3/(9-9sin^2theta)^(1/2)3cosd theta$

$\int \frac{9 cos θ}{{(9 [1 - {sin}^{2} θ])}^{\frac{1}{2}}} d θ$ $int (9costheta)/(9[1-sin^2theta])^(1/2)d theta$

$3 \int \frac{cos θ}{{({cos}^{2} θ)}^{\frac{1}{2}}} d θ$ $3int costheta/(cos^2theta)^(1/2)d theta$

$3 \int \frac{cos θ}{cos θ} d θ$ $3int costheta/costhetad theta$

$3 \int d θ$ $3int d theta$

Integrating we get $θ$ $theta$

Now $θ = arcsin (\frac{x}{3})$ $theta=arcsin(x/3)$

So we have $3 arcsin (\frac{x}{3})$ $3arcsin(x/3)$

Putting the two results together we have

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

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How do you find the integral of $\frac{2 x - 3}{{(9 - x^{2})}^{0.5}} d x$ $(2x-3)/((9-x^2)^0.5)dx$ ?

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Explanation:

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How do you find the integral of $\frac{2 x - 3}{{(9 - x^{2})}^{0.5}} d x$ $(2x-3)/((9-x^2)^0.5)dx$ ?