# Evaluate the integral (inverse trig)

shft600

## Homework Statement

Evaluate the following integral:
(int)1/sqrt(2x-x2)dx

## Homework Equations

d(arcsinx)/dx = 1/sqrt(1-x2)

## The Attempt at a Solution

I just need a good start in the right direction on this one so I can at least try it. I know the arcsin d/dx as above, but how do I make this work? Is it some kind of integration by parts or reduction? I have no idea how to go about dealing with the sqrt(2x-x2) mostly, or how it would get there from arcsin...

l'Hôpital
Try completing the square.

shft600
hm.. I'm trying that but still stuck, I get

int: 1/sqrt(x(2-x))dx, I can move the root up to the top easily enough by multiplying by sqrt(x(2-x))/sqrt(x(2-x)), but that doesn't help much. I found the formula relating a/sqrt(b+x2) to (a/sqrt(b))arctan(x/sqrt(b))+C, but that still doesn't help me with the 2x...

l'Hôpital
No no. -x^2 + 2x = -(x^2 - 2x) = -(ax + b)^2 + c. Find a, b, and c.

Once you do that, what happens if you let u = ax + b?

shft600
by that, it should be
-(x2-2x) = -((x-1)2-1), or a=1, b=-1, c=-1
so then, making u=x-1, du=1, it becomes

int{ du/sqrt(-u2-a2)

and that falls into the arccosh formula as:
cosh-1(-u/a)+c
or
cosh-1(-(x-1)/1)+c
cosh-1(1-x)+c, and that should be the final answer for the indefinate integral right? or can I just plug the -u in like that?

l'Hôpital
by that, it should be
-(x2-2x) = -((x-1)2-1), or a=1, b=-1, c=-1
so then, making u=x-1, du=1, it becomes

int{ du/sqrt(-u2-a2)

and that falls into the arccosh formula as:
cosh-1(-u/a)+c
or
cosh-1(-(x-1)/1)+c
cosh-1(1-x)+c, and that should be the final answer for the indefinate integral right? or can I just plug the -u in like that?

Be careful. You are saying -(x-1)^2 - 1 = 2x - x^2?

What if x = 1? Then, -1 = 2 - 1 = 1, which is wrong. It should be -(x-1)^2 + 1.
So if we let u = x - 1
So, we have,
$$\int \frac{1}{\sqrt{1 - u^2}} du$$
That should look more familiar.

shft600
ooh, my bad... so its equal to arcsin(x-1)+C !

l'Hôpital