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Could you please explain the following two questions relating to differential of a variable.

  1. In the method of integration by parts using substitution, we have $u = f(x)$, $v = g(x)$, $du = f'(x)dx$, and $dv = g'(x)dx$. Is it just a definition to assign those values to the differentials $du$ and $dv$ or is it based on some logics?

  2. How could the highlighted differentials $dt$ $du$ in the below text be derived? I tried to use the method in (1) above but it didn't work out as I got $sin\theta cos\theta (1 - r^2) + r(cos^2\theta - sin^2\theta)$.

enter image description here

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The point is that when you do substitution, you have $$ \int g(f(x))\,f'(x)\,dx=\int g(v)\,dv $$ by taking $v=f(x)$. Then the formula suggest that $f'(x)\,dx=dv$. The formula for integration by parts, that comes from the derivative of a product, is $$ \int f(x)\,g'(x)\,dx = f(x)g(x)-\int g(x)\,f'(x)\,dx. $$ With the above convention, and taking $u=g(x)$, you get $$ \int u\,dv=uv-\int v\,du. $$

The second case has nothing to do with the above. The situation is that now you have a double integral. When you do substitution in a double integral, taking $$ t=g(r,\theta),\ \ u=h(r,\theta), $$ the change of variable is given by $$ dt\,du=\begin{vmatrix} \frac{\partial g}{\partial r}&\frac{\partial g}{\partial \theta}\\ \frac{\partial h}{\partial r}&\frac{\partial h}{\partial \theta}\end{vmatrix}\,dr\,d\theta. $$ For the particular choice of polar coordinates in your example, the above determinant (called the Jacobian) is $r$.

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  • $\begingroup$ Wow! First time ever I've seen differential of a variable involving determinant. Thanks, Martin. Just a quick clarification: did you mean $\int g(f(x))\,f'(x)\,dx=\int g(v)\,dv$ instead of $\int g(f(x))\,f'(x)\,dx=\int f(v)\,dv$? $\endgroup$ – Nemo Jan 24 at 4:42
  • $\begingroup$ Indeed; edited. $\endgroup$ – Martin Argerami Jan 24 at 10:27
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Both t and u are functions of r and $ \theta $. Form the matrix J whose first row is $ \nabla t $ and whose second row is $ \nabla u $. Then when we change the variables, we have the formalism of replacing the symbol $ dt du $ by the symbol $ |\det(J)| dr d\theta $ You can easily check that $ |\det(J)$ $ is r.

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  • $\begingroup$ Thanks for your answer, @P. Lawrence. Without Martin spelling out, I don't think I could visualise your explanation. $\endgroup$ – Nemo Jan 24 at 4:50

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