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

• footmath
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 ?
footmath
please solve this integral : \[Integral]Sqrt[1/(u - 1) - 1/u] du

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 $u^2-u$
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:
yes . this integral at first was : $A=\int\sqrt{1+\sin^{2}x}\,dx$

so then that just equals cos(x)

cragar said:
so then that just equals cos(x)

sqrt{1-\sin^{2}x} equal cosx not sqrt{1+\sin^{2}x}

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 .

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

right but when you do the trig substitution it should help with that.

cragar said:
right but when you do the trig substitution it should help with that.

would you do it?

This is your problem. So far you have not shown any attempt yourself.

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 :$\int\sqrt{\frac{1}{f^{2}-f}}\,dt$

$\int\sqrt{\frac{1}{t^{2}-t}}dt =\int\frac{dt}{\sqrt{(t-\frac{1}{2})^{2}-\frac{1}{4}}}=ln\mid{t-\frac{1}{2}+\sqrt{t^{2}-t}}\mid+c$

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

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 ?

## 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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