Can you solve the funny integral with a square root of tangent?

[Gellmann]

[SOLVED] funny integral

hi everyone

its funny but all maths-software fail solving this "simple" integral

\int \sqrt{\tan x}\, dx

do you know another funny integrals?

 

[Timbuqtu]

Mathematica gives an answer:

\frac{-2 \tan ^{-1}\left(1-\sqrt{2} \tan ^{\frac{1}{2}}(x)\right)+2 \tan ^{-1}\left(\sqrt{2} \tan ^{\frac{1}{2}}(x)+1\right)+\log<br /> \left(-\tan (x)+\sqrt{2} \tan ^{\frac{1}{2}}(x)-1\right)-\log \left(\tan (x)+\sqrt{2} \tan ^{\frac{1}{2}}(x)+1\right)}{2<br /> \sqrt{2}}

 

[lurflurf]

\int \sqrt{\tan(x)}dx=\int\frac{\sec^2(x)}{1+\tan^2(x)}\sqrt{\tan(x)}dx=\int\frac{2\tan(x)}{1+\tan^2(x)}d\sqrt{\tan(x)}
so it hinges on the always fun
\int\frac{x^2}{1+x^4}dx

 

[Ali 2]

What is about this integral .. ?

\int_0^{\frac \pi 2} \ln ( \sin x ) . \ dx

Can the mathematical softwares, such as Maple amd Mathematica give you
the answer :- \frac \pi 2 \ln 2 ?

 

[hypermorphism]

What is about this integral .. ?

\int_0^{\pi/2} \ln ( \sin x ) . \ dx

Change ln to its integral form (so you get a double integral) and use change of variables on that form.

 

[Ali 2]

Change ln to its integral form (so you get a double integral) and use change of variables on that form.

Unfortunately, I edited my previous replay after you replied .. !

I wanted to say that the answer of this integral can't be obtained by Maple or Mathematica ..

Also , I solved the integral with a method different from your method .

 

[Madeline]

\int \sqrt{\tan(x)}dx=\int\frac{\sec^2(x)}{1+\tan^2(x)}\sqrt{\tan(x)}dx=\int\frac{2\tan(x)}{1+\tan^2(x)}d\sqrt{\tan(x)}
so it hinges on the always fun
\int\frac{x^2}{1+x^4}dx

Wow, I was looking up ways to figure out how to use the LaTeX graphics so that I type \int \sqrt{\tan(x)}dx and ask for help solving that. It's a funny coincidence that I stumbled into this thread. I'm a very lucky person.

Anyway... I can transform the integral into 2\int\frac{u^2}{1+u^4}du. I know what the antiderivative of that is is (I found it in a book of mathematical tables), but I don't know how to prove it. Do you happen to know how to find the antiderivative of \frac{u^2}{1+u^4}?

 

[amcavoy]

hi everyone
its funny but all maths-software fail solving this "simple" integral
\int \sqrt{\tan x}\, dx
do you know another funny integrals?

There was a long thread on this a while back. Can't find it though. Another strange one I've found (similar to one posted above) is:

\int\frac{1}{1+x^4}\,dx

You have to use that fact that x⁴+1=(x²+1)²-2x². And then use the difference of two squares. It gets messy.

Alex

 

[Tide]

hi everyone
its funny but all maths-software fail solving this "simple" integral
\int \sqrt{\tan x}\, dx
do you know another funny integrals?

You didn't try Derive 6! :)

 

[dextercioby]

hi everyone
its funny but all maths-software fail solving this "simple" integral
\int \sqrt{\tan x}\, dx
do you know another funny integrals?

Not all, my ancient version of Maple (5.3 i guess) gives

\int \sqrt{\tan x}dx=\frac{1}{2}\sqrt{2}\arctan \sqrt{2}\frac{\tan ^{\frac{1}{2}}x}{1-\tan x}-\frac{1}{2}\sqrt{2}\ln \frac{\tan x+\sqrt{2}\tan ^{\frac{1}{2}}x+1}{\sqrt{\left( 1+\tan ^{2}x\right) }} + C.

Daniel.

 

[Tide]

Gellman,

Incidentally, MuPAD also gives the correct result. What package were you using?

 

[Ali 2]

This a method to solve the integral ..
\int \frac { u^2 } { u^4 +1 } du =\frac 12 \int \frac { 2u^2 } { u^4 +1 } du = \frac 12 \int \frac { u^2 -1 } { u^4 +1 } du + \frac { u^2 +1 } { u^4 +1 } du
= \frac 12 \int \frac { 1 - \frac 1 {u^2} }{ u^2 + \frac 1{u^2} } du<br /> +\frac 12 \int \frac { 1 + \frac 1 {u^2} }{ u^2 + \frac 1{u^2} } du<br />
= \frac 12 \int \frac { 1 - \frac 1 {u^2} }{ \left (u + \frac 1u \right ) ^2 -1 } du<br /> +\frac 12 \int \frac { 1 + \frac 1 {u^2} }{ \left (u - \frac 1u \right)^2 +1 } du
=\frac 12 \int \frac { d \left (u + \frac 1u \right ) }{ \left (u + \frac 1u \right ) ^2 -1 } <br /> +\frac 12 \int \frac { d \left (u - \frac 1u \right ) }{ \left (u - \frac 1u \right)^2 +1 }

\mbox { In the first integral , make the subsitution :} v = u + \frac 1u
\mbox { and in the second integral , make the subsitution : } v = u - \frac 1u

The integrals become now simple , you can integrate them easily

 

[benorin]

There was a long thread on this a while back. Can't find it though. Another strange one I've found (similar to one posted above) is:
\int\frac{1}{1+x^4}\,dx
You have to use that fact that x⁴+1=(x²+1)²-2x². And then use the difference of two squares. It gets messy.
Alex

Yes, that fact indeed. Read attached gif, see what is meant.