From the product rule of differentiation, we can derive $duv = udv + vdu$. By integrating the both side, we can get $uv = \int{udv} + \int{vdu}$.
What is the sufficient condition that allows this equation, $uv = \int{udv} + \int{vdu}$?
I was confused by the following two equations.
If $u=x$, $v=y$, then $xy = \int{xdy} + \int{ydx}$. This is wrong.
If $u=\cos\theta$, $v=\sin\theta$, then $\cos\theta\sin\theta = \int{\cos{\theta}d\sin\theta} + \int{\sin\theta d \cos\theta}$. This is correct.
Because both $\left(x,y\right)$ and $\left(\cos\theta,\sin\theta\right)$ are linear independent, it seems that the linear independence is not the sufficient condition of the integration by parts. Did I right? Or I was confused vectors with functions?
Best Answer
In your first example ($u=x$ and $v=y$), the two functions are functions of different variables. But in your second example ($u=\cos\theta$ and $v=\sin\theta$), although they are independent, they are both functions of $\theta$, and so they are related in that way. In your first example, if it was instead $u=x$ and $v=y(x)$, then integration by parts would work.