improper integrals?

Question

integrate dx/ ((4*x^2)+(25)), the a=5 and b= positive infinity...i try to solve this a came up the result is in the trigonometric substitution the only problem how can I substitute the limits of 5 and infinity ??

integrate dx/(2x + 2)^1/2....can i use the general power formula to this question?

thanks

Answer 1

When working with improper integrals, rewrite them as limits of proper integrals.

In this case,
\int_{5}^{infinity} 1/(4x^2+25) dx=
\lim_{L -> infinity} \int_{5)^{L} 1/(4x^2+25) dx=
\lim_{L -> infinity} 1/10 (arctan(2L/5) - arctan (2))=
1/10 (pi/2 - arctan(2))

The second problem is easier. Let u=2x+2. Then, 1/2 du=dx so
\int 1/(2x+2)^(1/2) dx=
1/2 \int 1/u^(1/2) du=
1/2 \int u^(-1/2) du=
1/2 2 u^(1/2)+C=
(2x+2)^(1/2)+C

Notes:

1. \lim_{L \to \infinity} arctan(L)=pi/2

2. \int 1/(a^2+u^2) du=1/a arctan(u/a). In your case, u=2x so 1/2 du=dx and a=5 so
\int 1/(4x^2+25) dx=1/10 arctan(2x/5)+C

Good luck! Check to make sure I evaluated correctly! :)

Answer 2

The answer is in arctan, you have to do a bit of division and multiplication and manipulate the integral numerically.
Regarding the boundaries, note that arctan infinity is pi/2 rad.
(for proof consider y = arctan x => tan y =x, then differentiate and use the chain rule in the form 1/(dx/dy) = dy/dx)
For the second simply substitute(for sake of clarity, it'll come naturally to you after a while)
u = 2x+2
then du/dx = 2
then the integral becomes,
1/2. S u^-(1/2) du
where S stands for the integral sign.
Hope this helps!

Answer 3

Let I=integrate dx/ ((4*x^2)+(25)),
I=integrate 1/4dx/ {(x)^2+(5/2)^2}
Using the formula of integration
integrate of 1/(x^2+a^2)= (1/a) tan inverse of (x/a)
Here a =5/2

I=integrate (1/4)[dx/ {(x)^2+(5/2)^2}] =(1/4)*(2/5) tan inverse(2x/5) +C
I=1/10{tan inverse(infinity) - tan inverse(2) }
I=1/10[pi/2 - tan inverse(2)]

Answer 4

This is the complement of a parabola so the equation takes the form y^2 = 2paX,
dy = pax^-1/2
substitute infinity in x solve and subtract substituting 5 in x. Infinity can be pi/2 for x.

Answer 5

write in english man.


the first one can be done by using standard form of arctan. integral dx/ ((4*x^2)+(25)) = 1/10*arctan{(2x)/a}. Indefinite integral is evaluating between 5 and infinity (i think this is what you are asking). which is, 1/10*(arctan(oo)-arctan(5)). I guess this is right.
The second one involves log, with an adjustment of 1/2 - i'm not giving you the answer but maybe u or someone else can work out

Answer 6

Hi Dear Ice-Cream;

∫dx/ ((4*x^2)+(25 =
∫1/ ((4*x^2)+(25) dx = (1/10) tan^-1(2x/5) + c

[ (1/10) tan^-1(2(+∞) / 5) - [ (1/10) tan^-1(2(0)/5) ]

Good Luck Darling.