# Funny integral

1. Oct 2, 2005

### 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?

2. Oct 2, 2005

### Timbuqtu

$$\frac{-2 \tan ^{-1}\left(1-\sqrt{2} \tan ^{\frac{1}{2}}(x)\right)+2 \tan ^{-1}\left(\sqrt{2} \tan ^{\frac{1}{2}}(x)+1\right)+\log \left(-\tan (x)+\sqrt{2} \tan ^{\frac{1}{2}}(x)-1\right)-\log \left(\tan (x)+\sqrt{2} \tan ^{\frac{1}{2}}(x)+1\right)}{2 \sqrt{2}}$$

3. Oct 2, 2005

### 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
$$\int\frac{x^2}{1+x^4}dx$$

4. Oct 2, 2005

### Ali 2

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

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

Last edited: Oct 2, 2005
5. Oct 2, 2005

### hypermorphism

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

6. Oct 2, 2005

### Ali 2

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 .

Last edited: Oct 2, 2005
7. Oct 11, 2005

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 $$\frac{u^2}{1+u^4}$$?

Last edited: Oct 11, 2005
8. Oct 11, 2005

### amcavoy

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

9. Oct 12, 2005

### Tide

You didn't try Derive 6! :)

10. Oct 12, 2005

### dextercioby

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.

11. Oct 12, 2005

### Tide

Gellman,

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

12. Oct 18, 2005

### Ali 2

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

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

The integrals become now simple , you can integrate them easily

Last edited: Oct 18, 2005
13. Oct 20, 2005

### benorin

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

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