Solve Integral: $\sqrt{\frac{1}{u-1} - \frac{1}{u}}$

  • Thread starter footmath
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    Integral
In summary: I transformed$A=\int\sqrt{1+\sin^{2}x}\,dx$to$ \int\sqrt{\frac{t+1}{t^{2}-t}}\,dt$but how can I solve it ?
  • #1
footmath
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please solve this integral : \[Integral]Sqrt[1/(u - 1) - 1/u] du
 
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  • #2
have you tried a trig substitution or anything.
after looking at it a little bit combine the 2 factors under the root
and you get 1/(u(u-1))
now rewrite the bottom [itex] u^2-u [/itex]
now complete the square to rewrite the bottom as (u-.5)(u-.5)-.25
now we have something of the form x^2-a^2 . now do a u substitution first .
then do a trig substitution.
 
Last edited:
  • #3
yes . this integral at first was : $ A=\int\sqrt{1+\sin^{2}x}\,dx $
 
  • #4
so then that just equals cos(x)
 
  • #5
cragar said:
so then that just equals cos(x)

sqrt{1-\sin^{2}x} equal cosx not sqrt{1+\sin^{2}x}
 
  • #6
whoops , I edited my first post and I think it works combine that fraction then complete the square on the bottom. then do a trig substitution .
 
  • #7
cragar said:
whoops , I edited my first post and I think it works combine that fraction then complete the square on the bottom. then do a trig substitution .

you forget radical ? [Integral]Sqrt[1/(u - 1) - 1/u] du
 
  • #8
right but when you do the trig substitution it should help with that.
 
  • #9
cragar said:
right but when you do the trig substitution it should help with that.

would you do it?
 
  • #10
This is your problem. So far you have not shown any attempt yourself.
 
  • #11
HallsofIvy said:
This is your problem. So far you have not shown any attempt yourself.

I work this integral during one mouth . I am in grade two high school and my teacher gave me this integral : $ \int\sqrt{sin(u)}du $ and I transformed to : (sinx)^1/2=t => sinx=t^2 => x= arc sint^2
dx=2t/(1-t^4)^1/2
so $ \int\sqrt{sin(u)}du $=int (2t^2/(1-t^4)^1/2) if t=cosx we have : $ A=\int\sqrt{1+\sin^{2}t}\,dt $ if sqrt{1+\sin^{2}t}=f we have :$ \int\sqrt{\frac{1}{f^{2}-f}}\,dt $
 
  • #12
$ \int\sqrt{\frac{1}{t^{2}-t}}dt =\int\frac{dt}{\sqrt{(t-\frac{1}{2})^{2}-\frac{1}{4}}}=ln\mid{t-\frac{1}{2}+\sqrt{t^{2}-t}}\mid+c $
 
  • #13
footmath said:
yes . this integral at first was : $ A=\int\sqrt{1+\sin^{2}x}\,dx $

This integral cannot be done in terms of elementary functions. It can, however, be expressed in terms of a so-called "incomplete elliptic integral of the second kind". Are you sure you did not make an error in writing down the problem?

RGV
 
  • #14
Ray Vickson said:
This integral cannot be done in terms of elementary functions. It can, however, be expressed in terms of a so-called "incomplete elliptic integral of the second kind". Are you sure you did not make an error in writing down the problem?

RGV

no sorry.I transformed$A=\int\sqrt{1+\sin^{2}x}\,dx$to$ \int\sqrt{\frac{t+1}{t^{2}-t}}\,dt$
but how can I solve it ?
 

Related to Solve Integral: $\sqrt{\frac{1}{u-1} - \frac{1}{u}}$

1. What is an integral?

An integral is a mathematical concept that represents the area under a curve on a graph. It is used to find the total value of a continuously changing quantity, such as distance or volume, by breaking it down into infinitely small pieces.

2. How do you solve an integral?

To solve an integral, you must use a process called integration. This involves finding an antiderivative, which is the opposite of a derivative, and then evaluating the integral at specific upper and lower bounds.

3. What is a square root?

A square root is a mathematical operation that gives the number which, when multiplied by itself, gives the original number. In other words, it is the inverse of squaring a number.

4. Why is it important to solve integrals?

Solving integrals is important because it allows us to find the exact value of a quantity that is changing continuously. This is useful in many fields, including physics, engineering, and economics, where understanding the total value of a changing quantity is crucial.

5. What is the specific method for solving this integral: $\sqrt{\frac{1}{u-1} - \frac{1}{u}}$?

This integral can be solved using the substitution method. Letting $t = 1-u$, we can rewrite the integral as $\sqrt{\frac{1}{t} - \frac{1}{t+1}}$. Then, by finding the antiderivative and evaluating it at the bounds, we can solve for the final answer.

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