What is the solution for this integral?

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Homework Help Overview

The discussion revolves around the integral of the form \(\int \frac{\sin x \cos x}{\sin^4 x + x \cos^4 x} dx\). Participants are exploring various substitution methods and questioning the presence of the variable \(x\) in the denominator, which complicates the integral.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts a substitution with \(t = \cos x\) and raises concerns about the complexity of the resulting expression. Some participants question the validity of the substitutions made, particularly regarding the treatment of \(cos^4 x\) and its relation to \(arccos t\). Others express skepticism about the integral's solvability, noting that software like Maple and Mathematica struggle with it.

Discussion Status

There is an ongoing examination of the integral's structure, with participants sharing their experiences using computational tools. Some suggest that removing the variable \(x\) simplifies the integral significantly, while others note that integrals involving transcendental functions and variables are rarely expressible in terms of common special functions. The discussion reflects a mix of interpretations and approaches without reaching a consensus.

Contextual Notes

Participants are grappling with the implications of having a variable \(x\) in the denominator, which raises questions about the integral's solvability. There is also mention of differing outputs from computational tools, indicating potential discrepancies in results based on software versions.

twoflower
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Hi,

I've been having troubles solving this integral:

[tex] \int \frac{\sin x.\cos x}{\sin^4 x + x.\cos^4 x} dx[/tex]

Here's how I tried it:

[tex] t = \cos x [/tex]

[tex] dt = - \sin x dx[/tex]

[tex] dx = \frac{dt}{-\sin x}[/tex]

[tex] x = \arccos t[/tex]

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

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

[tex] \sin x = \sqrt{1-t^2}[/tex]

[tex] \int \frac{\sin x.\cos x}{\sin^4 x + x.\cos^4 x} dx = - \int \frac{\cos x}{\sin^4 x + x.\cos^4 x} . \left( -\sin x \right) dx = - \int \frac{t}{\left(1-t^2\right)^2 + t^4.\arccos t} dt[/tex]

Well, I don't know at all what to do with this now...

Does anybody have any idea?

PS: Why doesn't it insert newline, when I write "\\" in the LaTeX code?
 
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in the [itex]cos^4 x[/itex] in the denominator how did you manage to sibstitute arcSin t into it
you know cos x dx = dt but [itex]cos^4 x[/itex] is NOT arccos t

be careful of your substitutions!
 
Is this your integral?

[tex]\int \frac{\sin x\cos x}{\sin^{4}x+x\cos^{4}x} \ dx[/tex]...?

If so,then i got good news and bad news

Bad:Maple & Mathematica can't crack it.
Good:You can forget about it.


Daniel.
 
stunner5000pt said:
in the [itex]cos^4 x[/itex] in the denominator how did you manage to sibstitute arcSin t into it
you know cos x dx = dt but [itex]cos^4 x[/itex] is NOT arccos t

be careful of your substitutions!

I have no [itex]\arcsin x[/itex] nor [itex]\cos x dx = dt[/itex] there and I'm quite sure about the substitutions..
 
dextercioby said:
Is this your integral?

[tex]\int \frac{\sin x\cos x}{\sin^{4}x+x\cos^{4}x} \ dx[/tex]...?

If so,then i got good news and bad news

Bad:Maple & Mathematica can't crack it.
Good:You can forget about it.

Daniel.

I know, I tried Maple and Mathematica too..So I have some doubts about whether the x in the denominator should be there..Anyway, if I ever met such an integral (with variables outside trigonometric functions), is there any way to solve it?
 
Very good,but that still doesn't help.

Daniel.
 
If it isn't there,the integral is easy.But in general this type of integrals involving "x" & transcendental functions are very seldom expressible in terms of "common" special functions.

Daniel.
 
dextercioby said:
Very good,but that still doesn't help.

Daniel.

What did you reply with this to? :smile:
 
if there were no x in the denominator then it could be solved, it doesn't look as stubborn as the other integral

then [tex]\int \frac{\sin x\cos x}{\sin^{4}x+\cos^{4}x} \ dx = - \frac{1}{2} ArcTan (Cos(2x)) + C[/tex]
 
  • #10
dextercioby said:
If it isn't there,the integral is easy.But in general this type of integrals involving "x" & transcendental functions are very seldom expressible in terms of "common" special functions.

Daniel.

Yes, I tried it without the 'x' and it is easy. (BTW: Maple and Mathematica give different results. I got the Mathematica's one :smile: )
 
  • #11
stunner5000pt said:
if there were no x in the denominator then it could be solved, it doesn't look as stubborn as the other integral

then [tex]\int \frac{\sin x\cos x}{\sin^{4}x+\cos^{4}x} \ dx = - \frac{1}{2} ArcTan (Cos(2x)) + C[/tex]

Yes, that's the one I got too. Maple gives the same excepting the sign :)
 
  • #12
Nope,Maple gives the same answer

[tex]-\frac{1}{2} \arctan\left(\cos 2x\right) +C[/tex]

Daniel.
 
  • #13
dextercioby said:
Nope,Maple gives the same answer

[tex]-\frac{1}{2} \arctan\left(\cos 2x\right) +C[/tex]

Daniel.

int( (sin(x)*cos(x)) / ( (sin(x))^4 + (cos(x))^4 ),x); [ENTER]
[tex] 1/2\,\arctan \left( -1+2\, \left( \cos \left( x \right) \right) ^{2}<br /> \right)[/tex]
 
  • #14
My Maple is integrated in an ancient version of SWP.I guess it's less than 5.0 (i think,not too sure,though).So that's why it may be different from your answer.

I think your Maple is screwed up :-p

Daniel.
 
  • #15
Nope,the sign in the front of arctan is crucial.They're not the same function.

Daniel.
 
  • #16
dextercioby said:
My Maple is integrated in an ancient version of SWP.I guess it's less than 5.0 (i think,not too sure,though).So that's why it may be different from your answer.

I think your Maple is screwed up :-p

Daniel.

I have Maple 9.01 ;) Strange.
 
  • #17
haha, I just noticed the negative sign there and deleted my post :/

Didn't see it there before
 
  • #18
hmmmmmm

I have Maple 9.52 and it gives me the correct answer (with the minus sign)
 
Last edited:

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