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How to integrate dy on its own?

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


Posted from TSR Mobile
ddyy=1 \frac{d}{dy}y = 1
(if that's what you're asking for?)
So would it not just be a constant?

What's the full question anyway
Reply 3
Avatar for ps1265A
ps1265A
OP
14
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
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
Reply 5
Avatar for davros
davros
16
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.
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 ∫\displaystyle \int 1dy which becomes y

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