# Help with integration

1. Feb 5, 2005

### Yegor

A have to integrate x^4/Sqrt((1-x^2)^3)

I tried x = Sint, but it didn't help me.
What kind of substitution I should use?

2. Feb 5, 2005

### dextercioby

Yes,it can help u.4 times partial integration + watch oiut woth the signs and u'll get the result.

Daniel.

3. Feb 5, 2005

### Yegor

Hm.
I don't see even how to integrate dx/Sqrt((1-x^2)^3). (If i correctly understood Your idea)
Tehere should be any substitution

4. Feb 5, 2005

### dextercioby

Yes,the "sine" substitution will get u
$$\int \frac{\sin^{4}t}{\cos^{5}t} dt$$

,which can be solved via partial integration & simple fractions.

Daniel.

Last edited: Feb 5, 2005
5. Feb 5, 2005

### Yegor

No. "Sine" substitution gave me (Sint)^4/(Cost)^2
x=Sint
dx=Cost dt

Last edited: Feb 5, 2005
6. Feb 5, 2005

### dextercioby

What??
$$(1-x^{2})^{3}\rightarrow \cos^{6}t$$ agree??

$$dx\rightarrow \cos t \ dt$$

$$x^{4}\rightarrow \sin^{4}t$$

Make the proper ratio & u'll see that i was right.

Daniel.

Last edited: Feb 5, 2005
7. Feb 5, 2005

### arildno

$$x=sin(t)$$
$$\frac{x^{4}}{\sqrt{(1-x^{2})^{3}}}=\frac{\sin^{4}{t}}{\sqrt{\cos^{6}t}}=\frac{\sin^{4}t}{|\cos^{3}t|}$$

8. Feb 5, 2005

### dextercioby

UUUUUUUUUPS,i didn't see the sqrt...

Daniel;

9. Feb 5, 2005

### Yegor

Arildno, i think you have forgotten dx= Cost dt

10. Feb 5, 2005

### dextercioby

He hasn't he didn't deal with the "dx",just with the ratio...

Daniel.

11. Feb 5, 2005

### arildno

The one thing missing in your account, is the absolute value sign.
You should end up with:
$$\int{sign}(\cos{t})\frac{\sin^{4}t}{\cos^{2}t}dt$$

12. Feb 5, 2005

### dextercioby

Anyway,the advice with partial integration is still valid.U'll need the double angle formula,though.

Daniel.

13. Feb 5, 2005

### Yegor

I think I do not need partial integration. I can (Cost)^2=1-(Sint)^2 and then divide.
There are rather simple integrals.

But I realy don't understand why I must take into account sign(cost).
I have never seen anything similar

14. Feb 5, 2005

### arildno

Alternatively, you may rewrite this expression in terms of the tangent function.

15. Feb 5, 2005

### arildno

Well, you'll get wrong results if you don't.

It doesn't provide you with anymore trouble, though:
In regions when cos(t)>0, sign(cos(t))=1, whereas when cos(t)<0, sign(cos(t))=-1

16. Feb 5, 2005

### Yegor

I finally got the result. And it is correct! Thank You all.

17. Feb 5, 2005

### dextercioby

Yes,Arildo,things are much more simple that they were.U've forgotten where he started from
$$\sqrt{(1-x^{2})^{3}}$$

into reals has sense only for $-1<x<+1 [/tex] Making the substitution $$x\rightarrow \sin t$$ "t" must be in this interval and other (dictated by the periodicity of "sine") $$-\frac{\pi}{2}<t<+\frac{\pi}{2}$ where the cosine is strictly POSITIVE.No need for modulus... It can be proven that in all other real intervals where "sine t" obeys the inequality [tex] -1<\sin t<+1$$

the cosine is positive strictly.

Daniel.

18. Feb 5, 2005

### arildno

Oh, dear..the shame of it.. (goes back to my bucket)