# 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

3. Apr 11, 2005

### dextercioby

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

If so,then i got good news and bad news

Bad:Maple & Mathematica can't crack it.

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

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

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