# How do you find the integral of int sqrt(14x-x^2) dx?

##### 2 Answers
Sep 27, 2015

$\frac{49}{2} \arcsin \left(\frac{x - 7}{7}\right) + \frac{49}{4} \sin 2 \arcsin \left(\frac{x - 7}{7}\right) + C$

#### Explanation:

$\int \sqrt{14 x - {x}^{2}} \mathrm{dx} = \int \sqrt{49 - 49 + 14 x - {x}^{2}} \mathrm{dx} =$

$= \int \sqrt{49 - {\left(x - 7\right)}^{2}} \mathrm{dx} = 7 \int \sqrt{1 - {\left(\frac{x - 7}{7}\right)}^{2}} \mathrm{dx} = I$

$\frac{x - 7}{7} = \sin t \implies \mathrm{dx} = 7 \cos t \mathrm{dt}$

$I = 49 \int \sqrt{1 - {\sin}^{2} t} \cos t \mathrm{dt} = 49 \int \sqrt{{\cos}^{2} t} \cos t \mathrm{dt} =$

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

$I = \frac{49}{2} \arcsin \left(\frac{x - 7}{7}\right) + \frac{49}{4} \sin 2 \arcsin \left(\frac{x - 7}{7}\right) + C$

Refer to explanation

#### Explanation:

We first write the quantity under square root as follows

−x² + 14x = −(x² − 14x + 49 − 49) = −(x² − 14x + 49) − (−49) = 49 − (x − 7)²

Use a substitution:

x − 7 = 7 sinu =>dx = 7 cosu du

Hence the integral becomes

int sqrt(14x−x²) dx = int sqrt(49 − (x − 7)²) dx = int sqrt(49 − 49 sin²u) * 7 cosu du = ∫ 7 cosu * 7 cosu du = 49 ∫ cos²u du = 49 ∫ (1/2 + 1/2 cos(2u)) du = 49 (1/2 u + 1/4 sin(2u)) + c = 49 (1/2 u + 1/4 * 2 sinu cosu) +c = 49 (1/2 u + 1/2 sinu cosu) + c = 49 * 1/2 (u + sinu cosu) + c

Now we substitute back:

sinu = (x−7)/7
cosu = sqrt(1−sin²u) = sqrt(1−((x−7)/7)²) = sqrt((49−(x−7)²)/49) = sqrt(14x−x²)/7

Hence we have

49/2 (arcsin((x−7)/7) + (x−7)/7 * sqrt(14x−x²)/7) + c = 49/2 arcsin((x−7)/7) + 1/2 (x−7) sqrt(14x−x²) + c