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

1. Mar 10, 2005

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?

2. Mar 10, 2005

arildno

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

3. Mar 10, 2005

Zurtex

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

4. Mar 10, 2005

dextercioby

Are u sure it's
$$\int \sqrt{\cos x +\cos^{3}x} \ dx$$...?

Daniel.

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

Last edited: Mar 10, 2005
5. Mar 10, 2005

Zurtex

An obvious thought but are you sure it is not:

$$\int \sqrt{\cos (x) - {\cos^3 (x)}} dx$$

As this then becomes relatively simple.

6. Mar 10, 2005

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.

7. Mar 10, 2005

arildno

I think Zurtex has found the correct integrand ..

8. Mar 10, 2005

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.

9. Mar 10, 2005

dextercioby

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

Daniel.

10. Mar 10, 2005

arildno

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

11. Mar 10, 2005

dextercioby

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

Daniel.

12. Mar 10, 2005

arildno

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

13. Mar 10, 2005

dextercioby

For him it was a discovery...

He'll remember it as the integral that f***** up all mathematical software...:tongue2:

Daniel.

14. Mar 10, 2005

marlon

$$\sqrt{\cos x} \sqrt{1 + \cos^2x}$$
How about using t² = cos² and then calculate

$$\int \frac{-t \sqrt{1+t^2}}{\sqrt{1-t^2}} dt$$

Then integration by parts

$$\frac{-t}{\sqrt{1-t^2}} = d{\sqrt{1-t^2}}$$

For what it is worth

marlon

Last edited: Mar 10, 2005
15. Mar 10, 2005

dextercioby

$$|t|=|\cos x|$$ 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.

16. Mar 11, 2005

saltydog

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.

17. Mar 11, 2005

dextercioby

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

Daniel.

18. Mar 11, 2005

marlon

The integration by parts yields :

$$\int \sqrt{1+ t^2} d \sqrt{1-t^2}$$

$$\sqrt{1-t^2} \sqrt{1+t^2} - \int \frac{t \sqrt{1-t^2}}{\sqrt{1+t^2}} = \int \frac{-t \sqrt{1+t^2}}{\sqrt{1-t^2}} dt$$

Then add up the two integrals...
$$\int \frac{t \sqrt{1-t^2}}{\sqrt{1+t^2}} - \int \frac{t \sqrt{1+t^2}}{\sqrt{1-t^2}}$$

This sum yields

$$\int \frac{-2t^3 }{\sqrt {1-t^4}}$$

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

Last edited: Mar 11, 2005
19. Mar 11, 2005

dextercioby

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

Daniel.

20. Mar 11, 2005

marlon

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

marlon