trig integral?

Question

integral of sqrt(14x-x^2) dx

Answer 1

Well Dear Theresa It's your correct answer,

∫√(14x-x^2) dx =
( (√-(x-14)* x) ( (√x-14) (x-7)((√x) - 98 log( √x - 14 + √x ))) / (2√x-14) √x) +c

Ah my goodness SUCH A QUESTION...

Good Darling Theresa.

Answer 2

Hehehe. It's more than just a trig integral, it also has to be done by parts. The general form is
integral √(ax² + bx + c) dx. In your case, c = 0. a = -1, and b = 14. The integral is
(((2ax + b)/(4a))∙√(ax² + bx + c)) + ((4ac-b²)/(8a)) ∙integral
dx/(√(ax² + bx + c))
Then integral dx/(√(ax² + bx + c)) is
((-1)/√(-a)) arcsin((2ax+b)/√(b²-4ac))

Since c = 0 in your case, the arithmetic shouldn't be too bad ☺

You can also express that 2'nd integral in terms of logs or hyperbolics, but you asked for the trig (circular) function(s) in the answer.

Have fun ☺


Doug

Answer 3

14x-x^2=7^2-(x-7)^2

x-7=t
dx=dt
integral[ sqrt(7^2-t^2) ]

Try to complete by yourself.