# MATHS INTEGRATION, Wont let me post in calculus forum

MATHS INTEGRATION, Wont let me post in calculus forum :(

## Homework Statement

Integrate 1/(sqrt(2x-x^2)) between 0 and 2.

## The Attempt at a Solution

I have no idea what method to use to try and integrate it, ive played around with it for hours but it ends up not working out anyway, since The function is undefined at both 0 and 2 anyway.

so i have no idea??

i tried using simpsons rule to approximate but it doesnt get me close enough to the answer to justify my procedure

the answer is pi.
i just have no idea how to get it.

Sorry for posting in this forum but it wont let me start threads in the calculus and beyond one.

Thanks for any help!

## Homework Statement

Integrate 1/(sqrt(2x-x^2)) between 0 and 2.

## The Attempt at a Solution

I have no idea what method to use to try and integrate it, ive played around with it for hours but it ends up not working out anyway, since The function is undefined at both 0 and 2 anyway.

so i have no idea??

i tried using simpsons rule to approximate but it doesnt get me close enough to the answer to justify my procedure

the answer is pi.
i just have no idea how to get it.

Sorry for posting in this forum but it wont let me start threads in the calculus and beyond one.

Thanks for any help!
I was able to solve this with the substitution u=x-1, this results in an integral as follows

$$\int_{-1}^1{{du}\over{\sqrt{1-u^2}}}$$

It may or may not be obvious from here depending on how well you know your derivatives of inverse trig functions (e.g. inverse sine).

yesss makes sense, i can see it already, i tried something very similar to that, using substitutions and table of integrals, i had a resulting integral with inverse sin too! BUT i forgot to re-calculate the bounds.

Thanks a lot!! :)

butttt

that integral comes out to sin-1 (u/a).... with bounds -1 and 1 it gives sin-1(0) which is 0, and sin-1(-2) which is undefined

i think i still went wrong somewhere...

butttt

that integral comes out to sin-1 (u/a).... with bounds -1 and 1 it gives sin-1(0) which is 0, and sin-1(-2) which is undefined

i think i still went wrong somewhere...
Where did the "a" come from?

a just equals 1 anyway

let 1 = a so its 1/sqrt(a^2 - u^2) then from table of integrals its sin-1 u/a.. in this case its just sin-1 (u) since a = 1 anyway.

so then its [sin-1 (x-1)] between -1 and 1

a just equals 1 anyway

let 1 = a so its 1/sqrt(a^2 - u^2) then from table of integrals its sin-1 u/a.. in this case its just sin-1 (u) since a = 1 anyway.

so then its [sin-1 (x-1)] between -1 and 1
Yes, and you don't even need a table really. You just need to know the derivative of asin(u) with respect to u. (well, yeah that is what an integral table is, i guess)

yeah.. but then when you go to evaluate it you end up with asin(1-1)=asin(0)=0 and asin(-1-1)=asin(-2) which is undefined....

yeah.. but then when you go to evaluate it you end up with asin(1-1)=asin(0)=0 and asin(-1-1)=asin(-2) which is undefined....
No, check again. What is the sine of pi/2? What is the sine of -pi/2?

ahhh i worked it out... another silly mistake, im getting good at these...

Instead of subbing the bounds directly into u, i subbed the bounds into x-1 forgetting that id already re-calculated the bounds to allows for the substitution.

so it should be asin(1) and asin(-1) which yes comes out to 2 x pi/2 = pi.

thank you:)