- #1

**[SOLVED] solving integrals**

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.

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- Thread starter iceman
- Start date

- #1

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.

- #2

MathematicalPhysicist

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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. (-:

- #3

Lonewolf

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

- #4

MathematicalPhysicist

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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?

- #5

Lonewolf

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

- #6

MathematicalPhysicist

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i see, what is your way to this problem?

- #7

Lonewolf

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First, let's use a u-substitution.

Let u = x^{2} - 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*√(x^{2}-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/(y^{2}+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.

Let u = x

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*√(x

∫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/(y

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.

Last edited:

- #8

MathematicalPhysicist

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You have a mistake it should be x=(u+1)^1/2Originally posted by Lonewolf

First, let's use a u-substitution.

Let u = x^{2}- 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*√(x^{2}-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/(y^{2}+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.

- #9

Lonewolf

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

Last edited:

- #10

MathematicalPhysicist

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i don't understand how you got du/dx=2x

shouldnt it be: x^2-1/u+1?

shouldnt it be: x^2-1/u+1?

- #11

Lonewolf

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- #12

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

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