Integration problem?

Question

integral {dx/(sqrt 8x-x^2)}

Answer 1

Integral (dx / sqrt(8x - x^2) )

You have to solve this by completing the square.
Note that

8x - x^2
-x^2 + 8x
-(x^2 - 8x)

-(x^2 - 8x + 16) + 16
-(x - 4)^2 + 16

16 - (x - 4)^2

So our new integral is

Integral ( dx / sqrt(16 - (x - 4)^2) )

At this point, let u = x - 4. Then du = dx, so we have

Integral ( du / sqrt(16 - u^2) )

At this point, we use trig substitution. We let
u = 4sin(t), and du = 4cos(t) dt.

I'll leave it up to you to finish.

Answer 2

sin^-1 ((x/4)-1)

sin^-1 : sin inverse

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