Question

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

Answer 1

`49/2arcsin((x-7)/7)+49/4 sin 2arcsin((x-7)/7)+C`

Explanation:

`int sqrt(14x-x^2)dx= int sqrt(49-49+14x-x^2)dx=`

`=int sqrt(49-(x-7)^2)dx=7int sqrt(1-((x-7)/7)^2)dx=I`

`(x-7)/7=sint => dx=7costdt`

`I=49int sqrt(1-sin^2t)costdt=49int sqrt(cos^2t)costdt=`

`I=49 int cos^2tdt=49 int (1+cos2t)/2dt=49/2(t+1/2sin2t)+C`

`I=49/2arcsin((x-7)/7)+49/4 sin 2arcsin((x-7)/7)+C`

Answer 2

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#