Integral with sin() and cos()

1. Apr 11, 2005

twoflower

Hi,

I've been having troubles solving this integral:

$$\int \frac{\sin x.\cos x}{\sin^4 x + x.\cos^4 x} dx$$

Here's how I tried it:

$$t = \cos x$$

$$dt = - \sin x dx$$

$$dx = \frac{dt}{-\sin x}$$

$$x = \arccos t$$

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

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

$$\sin x = \sqrt{1-t^2}$$

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

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?

2. Apr 11, 2005

stunner5000pt

in the $cos^4 x$ in the denominator how did you manage to sibstitute arcSin t into it
you know cos x dx = dt but $cos^4 x$ is NOT arccos t

be careful of your substitutions!

3. Apr 11, 2005

dextercioby

Is this your integral?

$$\int \frac{\sin x\cos x}{\sin^{4}x+x\cos^{4}x} \ dx$$...?

If so,then i got good news and bad news

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

Daniel.

4. Apr 11, 2005

twoflower

I have no $\arcsin x$ nor $\cos x dx = dt$ there and I'm quite sure about the substitutions..

5. Apr 11, 2005

twoflower

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?

6. Apr 11, 2005

dextercioby

Very good,but that still doesn't help.

Daniel.

7. Apr 11, 2005

dextercioby

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.

8. Apr 11, 2005

twoflower

What did you reply with this to?

9. Apr 11, 2005

stunner5000pt

if there were no x in the denominator then it could be solved, it doesnt look as stubborn as the other integral

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

10. Apr 11, 2005

twoflower

Yes, I tried it without the 'x' and it is easy. (BTW: Maple and Mathematica give different results. I got the Mathematica's one )

11. Apr 11, 2005

twoflower

Yes, that's the one I got too. Maple gives the same excepting the sign :)

12. Apr 11, 2005

dextercioby

Nope,Maple gives the same answer

$$-\frac{1}{2} \arctan\left(\cos 2x\right) +C$$

Daniel.

13. Apr 11, 2005

twoflower

int( (sin(x)*cos(x)) / ( (sin(x))^4 + (cos(x))^4 ),x); [ENTER]
$$1/2\,\arctan \left( -1+2\, \left( \cos \left( x \right) \right) ^{2} \right)$$

14. Apr 11, 2005

dextercioby

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 :tongue2:

Daniel.

15. Apr 11, 2005

dextercioby

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

Daniel.

16. Apr 11, 2005

twoflower

I have Maple 9.01 ;) Strange.

17. Apr 11, 2005

Spectre5

haha, I just noticed the negative sign there and deleted my post :/

Didn't see it there before

18. Apr 11, 2005

Spectre5

hmmmmmm

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

Last edited: Apr 11, 2005