How do you find the integral of sqrt(13+12x-x^2)dx?

Feb 21, 2015

$\frac{49}{2} \arcsin \left(\frac{x - 6}{7}\right) + \frac{x - 6}{2} \sqrt{13 + 12 x - {x}^{2}} + c$.

$\int \sqrt{13 + 12 x - {x}^{2}} \mathrm{dx} = \int \sqrt{13 - \left({x}^{2} - 12 x\right)} \mathrm{dx} =$

$= \int \sqrt{13 - \left({x}^{2} - 12 x + 36 - 36\right)} \mathrm{dx} = \int \sqrt{13 + 36 - {\left(x - 6\right)}^{2}} \mathrm{dx} =$

$= \int \sqrt{49 - {\left(x - 6\right)}^{2}} \mathrm{dx} = \left(1\right)$.

Since

$x - 6 = 7 \sin t \Rightarrow x = 7 \sin t + 6 \Rightarrow \mathrm{dx} = 7 \cos t \mathrm{dt}$,

our integral becomes:

$\left(1\right) = \int \sqrt{49 - {\left(7 \sin t\right)}^{2}} \cdot 7 \cos t \mathrm{dt} =$

$= \int \sqrt{49 - 49 {\sin}^{2} t} \cdot 7 \cos t \mathrm{dt} =$

$= \int \sqrt{49 \left(1 - {\sin}^{2} t\right)} \cdot 7 \cos t \mathrm{dt} = \int 7 \sqrt{1 - {\sin}^{2} t} \cdot 7 \cos t \mathrm{dt} =$

$= 49 \int \sqrt{{\cos}^{2} t} \cdot \cos t \mathrm{dt} = 49 \int {\cos}^{2} t \mathrm{dt} = \left(2\right)$.

Since:

cos(alpha/2)=+-sqrt((1+cosalpha)/2,

our integral becomes:

$\left(2\right) = 49 \int \frac{1 + \cos 2 t}{2} \mathrm{dt} = \frac{49}{2} \left(\int \mathrm{dt} + \frac{1}{2} \int 2 \cos 2 t \mathrm{dt}\right) =$

$= \frac{49}{2} \left(t + \frac{1}{2} \sin 2 t\right) + c = \frac{49}{2} t + \frac{49}{2} \cdot \frac{1}{2} \cdot 2 \sin t \cos t + c =$

$= \frac{49}{2} t + \frac{49}{2} \sin t \cos t + c = \left(3\right)$.

Since

$\sin t = \frac{x - 6}{7} \Rightarrow t = \arcsin \left(\frac{x - 6}{7}\right)$

and

$\cos t = \sqrt{1 - {\sin}^{2} t} = \sqrt{1 - {\left(\frac{x - 6}{7}\right)}^{2}} =$

=sqrt(1-(x^2-12x+36)/49)=sqrt((49-x^2+12x-36)/49=

$= \frac{\sqrt{13 + 12 x - {x}^{2}}}{7}$,

our integral becomes:

$\left(3\right) = \frac{49}{2} \arcsin \left(\frac{x - 6}{7}\right) + \frac{49}{2} \cdot \frac{x - 6}{7} \cdot \frac{\sqrt{13 + 12 x - {x}^{2}}}{7} + c =$

$= \frac{49}{2} \arcsin \left(\frac{x - 6}{7}\right) + \frac{x - 6}{2} \sqrt{13 + 12 x - {x}^{2}} + c$.