# How do you calculate int6dx /sqrt[4-(x-1)^2]?

##### 1 Answer
Apr 3, 2015

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

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

To Evaluate : $6 \int \frac{1}{\sqrt{4 - {\left(x - 1\right)}^{2}}} \mathrm{dx}$

step1:

let $x - 1 = 2 \sin \theta \implies \frac{\mathrm{dx}}{d \theta} = 2 \cos \theta \implies \mathrm{dx} = 2 \cos \theta \cdot d \theta$

so that,

${\left(x - 1\right)}^{2} = 4 {\sin}^{2} \theta$

$4 - {\left(x - 1\right)}^{2} = 4 - 4 {\sin}^{2} \theta$

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

$\frac{1}{\sqrt{4 - {\left(x - 1\right)}^{2}}} = \frac{1}{2 \cos \theta}$

$\implies 6 \int \frac{1}{\sqrt{4 - {\left(x - 1\right)}^{2}}} \mathrm{dx} = 6 \int \frac{1}{2 \cos \theta} \cdot 2 \cos \theta \cdot d \theta$

$= 6 \int d \theta = 6 \cdot \theta + c$
Remember that $x - 1 = 2 \sin \theta \implies \theta = {\sin}^{-} 1 \left(\frac{x - 1}{2}\right)$

$\implies 6 \theta + c = 6 \cdot {\sin}^{-} 1 \left(\frac{x - 1}{2}\right) + c$