1. ## Quick Problem

I'm not entirely sure if this belongs in this forum, but it seemed too difficult for the traditional Calculus forums. Is there any way to integrate cos(x)=c*d2x/dt2, where x is a function of time and c is some arbitrary constant. Please and thank you.

2. >>cos(x)=c*d2x/dt2

I assume its more like :

cos(x)=c*d^2(x)/dt^2

=> 1/c * cos(x) = d^2(x)/dt^2

why not just integrate twice?

3. Originally Posted by Matter_Math
>>cos(x)=c*d2x/dt2

I assume its more like :

cos(x)=c*d^2(x)/dt^2

=> 1/c * cos(x) = d^2(x)/dt^2

why not just integrate twice?
If the question was $\frac{d^2 t}{dx^2} = \frac{1}{c} \cos (x)$ then yes, you could ntegrate twice.

But that's not the question. The question is $\frac{d^2 x}{dt^2} = \frac{1}{c} \cos (x)$, which is very different.

The usual method is to substitute $\frac{d^2 x}{dt^2} = \frac{d}{dx} \left(\frac{v^2}{2}\right)$ where $v = \frac{dx}{dt}$. Now integrate with respect to x to get v and then use v to get x.

4. The problem with that is that puts the cos(x) with the dt^2 for integration. It would be more like 1/cos(x)*d^2x=1/c*dt^2 which I can easily integrate once, but it becomes quite problematic the second time.

5. Originally Posted by FluidOfJustice
The problem with that is that puts the cos(x) with the dt^2 for integration. It would be more like 1/cos(x)*d^2x=1/c*dt^2 which I can easily integrate once, but it becomes quite problematic the second time.
Sorry but what you've posted makes absolutely no sense. I have told you how to solve the DE. Follow my advice, show your working and say where you're stuck if you still need help.

6. I was referring to the poster above you. Your solution helped enormously and I was able to solve the problem.