1. ## Integral Problem

Hi

can anyone help with this one?

(integral) dx/x^2 +2 dx

where the limits are b= srt2 and a=0

MathCad gives answer of 1/8pi2^0.5 but not sure how to get there by hand

2. Originally Posted by wannberocketscientist
Hi

can anyone help with this one?

(integral) dx/x^2 +2 dx

where the limits are b= srt2 and a=0

MathCad gives answer of 1/8pi2^0.5 but not sure how to get there by hand
$\displaystyle \int_0^{\sqrt{2}} \frac{dx}{x^2 + 2}$

Let $\displaystyle y = \frac{x}{\sqrt{2}} \implies dy = \frac{dx}{\sqrt{2}}$ thus
$\displaystyle \int_0^{\sqrt{2}} \frac{dx}{x^2 + 2} = \int_0^1 \frac{\sqrt{2}~dy}{2y^2 + 2}$

$\displaystyle = \frac{\sqrt{2}}{2} \int_0^1 \frac{dy}{y^2 + 1}$

Does this form give you any ideas about how to proceed?

-Dan

3. not sure do i now just apply :

f(x) = F(b) - F(a)?

4. Yes but you first have to get rid of that integral. Because you can't simply fill in an x value into that.

5. Originally Posted by wannberocketscientist
not sure do i now just apply :

f(x) = F(b) - F(a)?
In this case it would be $\displaystyle \int_0^1 f(y)~dy = F(1) - F(0)$, yes. Do you know what the function F is?

-Dan

6. got it:

2/srt2 * arctan(1) - arctan(0)= 0.5553603673

1/8*pi*2^0.5 = 0.553603673

So now that these match I need to express the answer in terms of pi - I am assuming that this means I have to turn the numerical value in terms of pi (e.g. pi/2)???

However this seems like an ugly solution so to speak as it would give:

5.66pi

doesnt sound right

7. $\displaystyle Arctan(x)\Big|^1_0 = \frac{\pi}{4} - 0$

Remember that tan(45 deg) = 1, so the tan of pi/4 is 1. So the Arctan of 1 is pi/4.

8. Originally Posted by wannberocketscientist
got it:

2/srt2 * arctan(1) - arctan(0)= 0.5553603673
And note that the original expression from my solution used $\displaystyle \frac{\sqrt{2}}{2}$, not $\displaystyle \frac{2}{\sqrt{2}}$

-Dan

9. that was a typo sorry!

10. Dan,

In your first post here you change the expression to put srt2 and 2 into the fraction. Could you explain where the 2 comes from that is placed in front of the x2?

Thanks.

11. Well he did $\displaystyle y = \frac{x}{\sqrt{2}}$. So basically he took that and solved for x to get $\displaystyle x = \sqrt{2}y$. Then he needed to square that because the question has x^2.

I would done it in this order though:

$\displaystyle \int \frac{dx}{x^2 + 2} = \int \frac{dx}{2((\tfrac{x}{\sqrt{2}})^2 + 1)}$

Then you can see that the substitution of y= x/sqrt(2) is quite simple.

12. Originally Posted by wannberocketscientist
Dan,

In your first post here you change the expression to put srt2 and 2 into the fraction. Could you explain where the 2 comes from that is placed in front of the x2?

Thanks.
To be honest, when I did the problem the first time I put $\displaystyle x^2 = 2y^2$ because that's what I saw to be useful. But I thought that putting the substitution in this form would be more confusing than the one that I chose. Perhaps not.

-Dan