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**numen****Member**- Registered: 2006-05-03
- Posts: 115

the problem:

y' cos(x) + 2y sin(x) = ([sin(x)]^3)/(cos(x))

I know I could divide by cos(x) and calculate an integration factor, which I think would be 1/[cos(x)]^2 ?

Then I get: D(y/([cos(x)]^2)) = ([sin(x)]^3)/([cos(x)]^3)

Thus, the integral follows: ∫ ([sin(x)]^3)/([cos(x)]^4) dx, which I'm not sure how to solve...

Any ideas? I'm kinda stuck with this one, might be something wrong somewhere. My head just keeps spinning around

Bang postponed. Not big enough. Reboot.

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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

http://www.mathsisfun.com/forum/viewtop … 635#p32635

Using that formula, I get:

This ugly equation comes out to be:

"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."

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**numen****Member**- Registered: 2006-05-03
- Posts: 115

I've learned to not use formulas straight off like that, I learn nothing from it. Kinda ugly, yeah. But thanks anyway for showing that formula, might get in handy, though I preferably do everything from scratch.

Do you know, though, how to continue from where I left? Anyone else? I'll think about it some more, I simply can't sleep until I've figured it out =P

btw, is that LaTeX? How do I put it in here?

Bang postponed. Not big enough. Reboot.

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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

Since the equation is not seperable and not homogenous, I don't believe there is any other way to solve it.

"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."

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**ganesh****Moderator**- Registered: 2005-06-28
- Posts: 18,124

numen.

Here is a link on how to use LaTeX, in this forum or any other place.

**The divergent series are the invention of the devil, and its a shame to base on them any demonstration whatsoever - Neils Henrik Abel**.**Knowledge is of two kinds. We know a subject ourselves, or we know where we can find information upon it - Samuel Johnson.**

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**George,Y****Member**- Registered: 2006-03-12
- Posts: 1,306

I've written the proof or the derivation in post 3, following Ricky's fomula, so you can check my procedure and simply duplicate it before showing the formula. And that will make your solution complete.

*Last edited by George,Y (2006-05-04 15:52:44)*

**X'(y-Xβ)=0**

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**numen****Member**- Registered: 2006-05-03
- Posts: 115

It's solved now. Thanks George, even though it's not what I was looking for... again.

Anyway, the (easier) solution is y = 1/3cos(x)-cos(x)+C[cos(x)]^2. You get it by rewriting the integral I ended up with by using trigonometry rules and then use variable substitution to simplify the process.

Now I can go to sleep again :] thanks ganesh for that link, will have a look.

Bang postponed. Not big enough. Reboot.

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**George,Y****Member**- Registered: 2006-03-12
- Posts: 1,306

Hey, I know how to solve it now, too!

The keys are

a) the derivative of cos²(x) is -2cos(x)sin(x)

b)

Your question is so inspiring! I used to think this kinda trig and derivative mix is my last-to-solve.

*Last edited by George,Y (2006-05-05 14:04:53)*

**X'(y-Xβ)=0**

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**yttrium88****Member**- Registered: 2005-12-01
- Posts: 20

As far as that first integral, i would just replace (sin x)^3 with (1-(cos x)^2)(sin x) and sub in with u=cos x

du=-sinx dx

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