How to integrate sqrt( cos(x) + cos(x)^3)

[Yegor]

I need to integrate Sqrt[Cos(x)+(Cos(x))^3]
Mathematica gives answer with AppelF, I've never heard about it. This excercize was given in my book, so there must be answer in elementary functions. I tryed 1) tg(x/2)=t and 2) Cosx=t , but both didn't help me. Give me please any advice.
What does AppelF mean?

 

[arildno]

In all probability, "Appel function".
You might try to find out more about it on wikipedia.

 

[Zurtex]

I would suggest you double check the question, mathematica will just plain not compute that for me.

 

[dextercioby]

Are u sure it's
[tex] \int \sqrt{\cos x +\cos^{3}x} \ dx [/tex]...?

Daniel.

EDIT:My Maple is helpless.It reminds me of another integral...
[tex] \int e^{\sin x} dx [/tex]

 

[Zurtex]

An obvious thought but are you sure it is not:

[tex]\int \sqrt{\cos (x) - {\cos^3 (x)}} dx[/tex]

As this then becomes relatively simple.

 

[Yegor]

Yes, Daniel, You are right. In fact it was definite integral, but I don't remember borders x1 and x2. I think it doesn't matter.

 

[arildno]

I think Zurtex has found the correct integrand ..

 

[dextercioby]

It doesn't.Wolfram Mathematica's Integrator is helpless,too.

I'll look it up in G&R,though i doubt it is there.

Daniel.

 

[dextercioby]

U can't apply the FTC,because you can't find the antiderivative.

Daniel.

 

[arildno]

U can't apply the FTC,because you can't find the antiderivative.

Daniel.

Eeh, assuming Zurtex is right, the anti-derivative [tex]\int\sqrt{\cos{x}-\cos^{3}x}dx=\int|\sin{x}|\sqrt{\cos{x}}dx[/tex] is easily expressed in terms of elementary functions.

 

[dextercioby]

Yes,i was OBVIOUSLY referring to the OP's problem,not to an invented one...

Daniel.

 

[arildno]

Isn't the OP's problem invented??
Surely, it wasn't discovered..,

 

[dextercioby]

For him it was a discovery...

He'll remember it as the integral that f***** up all mathematical software...

Daniel.

 

[marlon]

[tex]\sqrt{\cos x} \sqrt{1 + \cos^2x}[/tex]
How about using t² = cos² and then calculate


[tex]\int \frac{-t \sqrt{1+t^2}}{\sqrt{1-t^2}} dt[/tex]

Then integration by parts

[tex]\frac{-t}{\sqrt{1-t^2}} = d{\sqrt{1-t^2}}[/tex]

For what it is worth

marlon

 

[dextercioby]

Your substitution is something like

[tex]|t|=|\cos x| [/tex] which is very difficult to manipulate.

Besides,the function under the sq.root can take both positive & negative values.And that's why a substitution must be made with care...

Daniel.

 

[saltydog]

I need to integrate Sqrt[Cos(x)+(Cos(x))^3]
Mathematica gives answer with AppelF, I've never heard about it. This excercize was given in my book, so there must be answer in elementary functions. I tryed 1) tg(x/2)=t and 2) Cosx=t , but both didn't help me. Give me please any advice.
What does AppelF mean?

Yegor, can you report the solution technique? What section in the Calculus text did this integral appear? That might give a clue as to how the text wished it to be solved.

 

[dextercioby]

Salty,i'm 99% sure that it wasn't that integral he was suppossed to solve...

Daniel.

 

[marlon]

[tex]\sqrt{\cos x} \sqrt{1 + \cos^2x}[/tex]
How about using t² = cos² and then calculate


[tex]\int \frac{-t \sqrt{1+t^2}}{\sqrt{1-t^2}} dt[/tex]

Then integration by parts

[tex]\frac{-t}{\sqrt{1-t^2}} = d{\sqrt{1-t^2}}[/tex]

For what it is worth

marlon

The integration by parts yields :

[tex] \int \sqrt{1+ t^2} d \sqrt{1-t^2}[/tex]

[tex]\sqrt{1-t^2} \sqrt{1+t^2} - \int \frac{t \sqrt{1-t^2}}{\sqrt{1+t^2}} = \int \frac{-t \sqrt{1+t^2}}{\sqrt{1-t^2}} dt[/tex]


Then add up the two integrals...
[tex] \int \frac{t \sqrt{1-t^2}}{\sqrt{1+t^2}} - \int \frac{t \sqrt{1+t^2}}{\sqrt{1-t^2}}[/tex]

This sum yields

[tex]\int \frac{-2t^3 }{\sqrt {1-t^4}}[/tex]

This is easy to solve

Indeed keep in mind that with the substitutions you need to be sure the sign is + otherwise they are invalid. Now you can incorporate negative numbers by writing a minus and this works, i checked it out

marlon

 

[dextercioby]

Marlon,it's chasing your tail...You get nowhere,trust me.

Daniel.

 

[marlon]

Marlon,it's chasing your tail...You get nowhere,trust me.

Daniel.

dexter don't be so jalous... Prove it...Prove me wrong...

marlon

 

[dextercioby]

If you call the initial integral I and the second (after applying partial integration) by J,u have that

[tex] I+J=\sqrt{1-t^{4}} [/tex] (1)

And let's do what u prescribed.Reevaluate this sum starting with the initial forms of I & J,namely:

[tex] I+J=-2\int\frac{t^{3}dt}{\sqrt{1-t^{4}}} =\sqrt{1-t^{4}} [/tex](2)

,which (normally) concides with (1).This is what i call chasing your tail.

Daniel.

EDIT:Sure,the constants of integration can be put at the end of the calculations.

 

[marlon]

hmmm, forget about it...the solution i gave does not lead to anything indeed

Are we sure we cannot solve this integral ?

marlon

 

[marlon]

How about this. let's take the "original" integral :

[tex]\int \frac{-t \sqrt{1+t^2}}{\sqrt{1-t^2}} dt[/tex]

now multiply this fraction by [tex]\frac{\sqrt{1+t^2}}{\sqrt{1+t^2}}[/tex]

we get
[tex]\int \frac{-t}{\sqrt{1-t^4}} - \int \frac{t^3}{\sqrt{1-t^4}}[/tex]

The first integral can be solved by a substitution t²=cosx

Thus : 2tdt = -sinx dx

I get

[tex]1/2 \int \frac {sinx dx}{\sqrt{1-cosx^2}}[/tex]

marlon

 

[dextercioby]

That is not the "original" integral.The original integral is
[tex] \int\sqrt{\cos x}\sqrt{1+cos^{2}x} \ dx [/tex]

And your first substitution is wrongly made...

Don't bother to edit that post.It would be useless.You haven't proven anyhting...

Daniel.