How do you calculate $\int 6 \frac{d x}{\sqrt{4 - {(x - 1)}^{2}}}$ $int6dx /sqrt[4-(x-1)^2]$ ?

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How do you calculate $\int 6 \frac{d x}{\sqrt{4 - {(x - 1)}^{2}}}$ $int6dx /sqrt[4-(x-1)^2]$ ?

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Answer 1 · 2015-04-03T16:42:43

Answer is:
$6 {sin}^{- 1} (\frac{x - 1}{2}) + c$ $6sin^-1((x - 1)/2) + c$

To me the me the easiest way to evaluate this is to use a substitution that takes $sin x$ $sinx$
Because $\int \frac{1}{\sqrt{1 - x^{2}}} d x = {sin}^{- 1} (x) + c$ $int1/(sqrt(1 - x^2))dx = sin^-1(x) + c$
Right?

**To Evaluate : ** $6 \int \frac{1}{\sqrt{4 - {(x - 1)}^{2}}} d x$ $6int1/(sqrt(4 - (x - 1)^2))dx$

step1:

let $x - 1 = 2 sin θ \Rightarrow \frac{d x}{d θ} = 2 cos θ \Rightarrow d x = 2 cos θ \cdot d θ$ $x - 1 = 2sintheta => (dx)/(d theta) = 2cos theta => dx = 2costheta*d theta$

so that,

${(x - 1)}^{2} = 4 {sin}^{2} θ$ $(x - 1)^2 = 4sin^2theta$

$4 - {(x - 1)}^{2} = 4 - 4 {sin}^{2} θ$ $4 - (x - 1)^2 = 4 - 4sin^2theta$

$\sqrt{4 - {(x - 1)}^{2}} = \sqrt{4 - 4 {sin}^{2} (θ)} = \sqrt{4 (1 - {sin}^{2} θ)} = \sqrt{4 {cos}^{2} θ}$ $sqrt(4 - (x - 1)^2) = sqrt(4 - 4sin^2(theta)) = sqrt(4(1 - sin^2theta)) = sqrt(4cos^2theta)$
$= 2 cos θ$ $= 2costheta$

$\frac{1}{\sqrt{4 - {(x - 1)}^{2}}} = \frac{1}{2 cos θ}$ $1/(sqrt(4 - (x - 1)^2)) = 1/(2costheta)$

$\Rightarrow 6 \int \frac{1}{\sqrt{4 - {(x - 1)}^{2}}} d x = 6 \int \frac{1}{2 cos θ} \cdot 2 cos θ \cdot d θ$ $=> 6int1/(sqrt(4 - (x - 1)^2))dx = 6int1/(2costheta)*2costheta*d theta$

$= 6 \int d θ = 6 \cdot θ + c$ $= 6intd theta = 6*theta + c$
Remember that $x - 1 = 2 sin θ \Rightarrow θ = {sin}^{- 1} (\frac{x - 1}{2})$ $x - 1 = 2sintheta => theta = sin^-1((x - 1)/2)$

$\Rightarrow 6 θ + c = 6 \cdot {sin}^{- 1} (\frac{x - 1}{2}) + c$ $=> 6theta + c = 6*sin^-1((x - 1)/2) + c$

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How do you calculate $\int 6 \frac{d x}{\sqrt{4 - {(x - 1)}^{2}}}$ $int6dx /sqrt[4-(x-1)^2]$ ?

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