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# How to integrate dy on its own? watch

1. Confused about this one because is it the same as the integral 0 dy

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2. (if that's what you're asking for?)
3. So would it not just be a constant?

What's the full question anyway
4. (Original post by L'Evil Fish)
So would it not just be a constant?

What's the full question anyway
dy/dx = sinxcos^2 x
I have to integrate this using a particular formula 1/g(y) = f(x) both sides integrated

So I end up with integral dy = integral sinxcos^2 x

But the mark scheme integrates this to y = -cos^3 x / 3.... I don't understand why it's "y"

Posted from TSR Mobile
5. (Original post by ps1265A)
dy/dx = sinxcos^2 x
I have to integrate this using a particular formula 1/g(y) = f(x) both sides integrated

So I end up with integral dy = integral sinxcos^2 x

But the mark scheme integrates this to y = -cos^3 x / 3.... I don't understand why it's "y"

Posted from TSR Mobile
Oh because d/dy will just integrate to y

I thought you were differentiating constant
6. (Original post by ps1265A)
dy/dx = sinxcos^2 x
I have to integrate this using a particular formula 1/g(y) = f(x) both sides integrated

So I end up with integral dy = integral sinxcos^2 x

But the mark scheme integrates this to y = -cos^3 x / 3.... I don't understand why it's "y"

Posted from TSR Mobile
Why do you have to use that particular formula involving "g(y)"?

You know that integration reverses differentiation, so the integral of dy/dx w.r.t.x is just y itself (or y + c if you want to be pedantic). So your only problem is to integrate sinxcos^2x w.r.t.x which is clearly -(cos^3x)/3 either by inspection or direct substitution.

So your final answer is just y = -(cos^3x)/3 + k where k is some constant.
7. (Original post by ps1265A)
dy/dx = sinxcos^2 x
I have to integrate this using a particular formula 1/g(y) = f(x) both sides integrated

So I end up with integral dy = integral sinxcos^2 x

But the mark scheme integrates this to y = -cos^3 x / 3.... I don't understand why it's "y"

Posted from TSR Mobile
Separation of variables is the method

You have 1 dy/dx

So, after separation of variables you have 1dy which becomes y

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