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

[footmath]

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

 

[cragar]

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.

 

[footmath]

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

 

[cragar]

so then that just equals cos(x)

 

[footmath]

so then that just equals cos(x)

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

 

[cragar]

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 .

 

[footmath]

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

 

[cragar]

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

 

[footmath]

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

would you do it?

 

[HallsofIvy]

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

 

[footmath]

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 $

 

[footmath]

$ \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 $

 

[Ray Vickson]

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

 

[footmath]

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 ?