Consider the following iterated integral \begin{gather*} \int_{0}^{1}\int_{0}^{\pi}y\sin(xy)d xd y.\tag{1} \end{gather*} It is standard to evaluate (1) as follows. \begin{align*} \int_{0}^{1}\int_{0}^{\pi}y\sin(xy)d xd y&=\int_{0}^{1}\int_{0}^{\pi}\frac{\partial}{\partial x}\big(-\cos(xy)\big)d xd y\\ &=\int_{0}^{1}\big(-\cos(xy)\big)\Big|_{x=0}^{x=\pi} dy=\int_{0}^{1}\big(1-\cos(\pi y)\big) dy\\ &=\left(y-\frac{\sin(\pi y)}{\pi}\right)\bigg|_0^1=1. \end{align*} My question is: Can I write the inner integral $\int_{0}^{\pi}y\sin(xy) d x$ as \begin{gather*} \int_{0}^{\pi}y\sin(xy) d x=\int_0^{\pi} d_x \big(-\cos(xy)\big), \end{gather*} so that the iterated integral is \begin{align*} \int_{0}^{1}\int_{0}^{\pi}y\sin(xy) d x dy=\int_{0}^{1}\int_{0}^{\pi}d_x \big(-\cos(xy)\big) dy?\tag{2} \end{align*} Here the notation $d_x$ in (2) means that the operation of differentiation is applied to just $x$ variable, holding the $y$ as constant. That is, $d_x\big(-\cos(xy)\big)$ means \begin{gather*} \frac{\partial}{\partial x}\big(-\cos(xy)\big)d x. \end{gather*} If we drop off the subscription $x$ in the notation $d_x,$ then \begin{gather*} d\big(-\cos(xy)\big)=\sin(xy)\big(yd x+xd y\big), \end{gather*} which is the (total) differential of $-\cos(xy).$ But this is not what I want.
Similarly, $d_y f(x,y)$ means \begin{gather*} d_yf(x,y)=\frac{\partial f(x,y)}{\partial y}d y. \end{gather*} Why I write $d_x$ and/or $d_y$ in the iterated integral? Because with this notation, the following calculation (that is, using push forward substitution and integration by parts) is easy to understand: \begin{align*} &\int_{0}^{1}\int_{0}^{\pi}y\sin(xy)d xd y=\int_{0}^{\pi}\int_{0}^{1}y\sin(xy) dy dx =\int_{0}^{\pi}\int_{0}^{1} \frac{y}{x}d_y \big(-\cos(xy)\big)d x\\ =&\int_{0}^{\pi}\left(\frac{y}{x}\big(-\cos(xy)\big)\Big|_{y=0}^{y=1}-\int_{0}^{1}\big(-\cos(xy)\big) d_y \Big(\frac{y}{x}\Big)\right)d x\\ =&\int_{0}^{\pi}\left(\frac{-\cos(x)}{x}+\int_{0}^{1}\frac{\cos(xy)}{x}dy\right)d x=\int_{0}^{\pi}\left(\frac{-\cos(x)}{x}+\int_{0}^{1}\frac{\cos(xy)}{x^2}d_y(xy)\right)d x\\ =&\int_{0}^{\pi}\left(\frac{-\cos(x)}{x}+\int_{0}^{1}\frac{1}{x^2}d _y\big(\sin(xy)\big)\right)d x =\int_{0}^{\pi}\left(\frac{-\cos(x)}{x}+\frac{\sin(xy)}{x^2}\Big|_{y=0}^{y=1}\right)d x\\ =&\int_{0}^{\pi}\left(\frac{-\cos(x)}{x}+\frac{\sin(x)}{x^2}\right)d x=\frac{-\sin(x)}{x}\bigg|_{0}^{\pi}=1. \end{align*} If without $d_x$ or $d_y$ notation, the above calculation would be like this (just as in standard textbooks): \begin{align*} &\int_{0}^{1}\int_{0}^{\pi}y\sin(xy)d xd y=\int_{0}^{\pi}\int_{0}^{1}y\sin(xy) dy dx\\ =&\int_{0}^{\pi}\left(\frac{-y\cos(xy)}{x}+\frac{\sin(xy)}{x^2}\right)\bigg|_{y=0}^{y=1}d x\tag{3}\\ =&\frac{-\sin(x)}{x}\bigg|_{0}^{\pi}=1. \end{align*} But the result in (3) is to easy to understand for beginners.