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
got it:
2/srt2 * arctan(1) - arctan(0)= 0.5553603673
MathCad Solution :
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
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.