[SOLVED] solving integrals

[iceman]

Hello, I need your help please?


It might seem obvious to you but I can't seem to get my head around how to solve this one.

int[1/(x*sqrt(x^2-1))]dx , from lower boundary =1, upper boundary = infinity.

Thanks indvance.

[loop quantum gravity]

Will need someone else to verify my answer but here goes nothing.
int[1/(x*sqrt(x^2-1))]dx , from lower boundary =1, upper boundary = infinity
the answer is: (replacing infinity with t) lim t->infinity[1/2ln(t^4-t^2)-1/2ln0]=
-1/2ln0, ln0 is undefined from this there is no answer.

Hope my answer is correct. (-:

[Lonewolf]

I got Pi/2, and the integral to be arctan(√(x2-1). How did you approach this, LQG?

[loop quantum gravity]

y=x*sqrt(x^2-1)
int[1/(x*sqrt(x^2-1))]dx
int[1/y]dx
as you know the inegral of 1/y=lny+c

what's wrong with it?
[:)]

[Lonewolf]

∫1/ydy = ln(y)

You integrated w.r.t y, when you should have integrated w.r.t x. Try a substitution.

[loop quantum gravity]

i see, what is your way to this problem?

[Lonewolf]

First, let's use a u-substitution.

Let u = x2 - 1
x = √(u+1)

(u cannot be less than 0, so we can ignore the negative case)

Then, du/dx = 2x, so du = 1/2x

The integral ∫1/(x*√(x2-1)dx now becomes

∫1/(2u+2)√(u)du which simplifies to 1/2∫du/sqrt(u)(u+1)

Now, we notice something...if y = arctan(z), dy/dz = 1/(y2+1). So, if y = arctan(√z), dy/dz = 1/(2√z(z+1)), which looks exactly like our integral. Thus we now know 1/2∫du/sqrt(u)(u+1) = arctan(√u). (Can't think of a method that doesn't use foresight just yet).

Now, we have the problem of actually evaluating this from x = 1 to infinity. We shan't bother substituting u in terms of x. Instead, we work out the values of u at x=1 and infinity (0 and infinity respectively).

lim t-> infinity (arctan(t)) - arctan(0) = Pi/2 - 0 = Pi/2, since the tangent function approaches infinity as the angle approaches Pi/2 radians.

[loop quantum gravity]

Originally posted by Lonewolf
First, let's use a u-substitution.

Let u = x2 - 1
x = √(u-1)

(u cannot be less than 0, so we can ignore the negative case)

Then, du/dx = 2x, so du = 1/2x

The integral ∫1/(x*√(x2-1)dx now becomes

∫1/(2u+2)√(u)du which simplifies to 1/2∫du/sqrt(u)(u+1)

Now, we notice something...if y = arctan(z), dy/dz = 1/(y2+1). So, if y = arctan(√z), dy/dz = 1/(2√z(z+1)), which looks exactly like our integral. Thus we now know 1/2∫du/sqrt(u)(u+1) = arctan(√u). (Can't think of a method that doesn't use foresight just yet).

Now, we have the problem of actually evaluating this from x = 1 to infinity. We shan't bother substituting u in terms of x. Instead, we work out the values of u at x=1 and infinity (0 and infinity respectively).

lim t-> infinity (arctan(t)) - arctan(0) = Pi/2 - 0 = Pi/2, since the tangent function approaches infinity as the angle approaches Pi/2 radians.
You have a mistake it should be x=(u+1)^1/2

[Lonewolf]

Yes, it should. Well spotted. Just a typo though, the result holds. I'll go edit it.

[loop quantum gravity]

i dont understand how you got du/dx=2x
shouldnt it be: x^2-1/u+1?

[Lonewolf]

The substitution u = x2 - 1 was used. Differentiate this w.r.t. x to get du/dx. Seperate the differentials to get du in terms of dx. We need this to integrate in terms of du.

[suffian]

this integral is made easiest by recognizing it's just arcsec|x| (abs bars or not depending on how you define arcsec)

arcsec(inf) - arcsec(1) = pi/2 - 0 = pi/2