Name of the rule allowing the exchanging $\sin$ and $\cos$ in integrals with limits $0$ and $\pi/2$? As in $0$ to $\frac{\pi}{2}$ limits the area under curve of $\sin \theta$ and $\cos \theta$ are same, so in integration if the limits are from $0$ to $\frac{\pi}{2}$ we can replace $\sin \theta$ with $\cos \theta$ and vice versa. Example-
\begin{align*}
   \int\limits_{0}^{\frac{\pi}{2}} \frac{\sin^3x-\cos x}{\cos^3x-\sin x} dx &=\int\limits_{0}^{\frac{\pi}{2}} \frac{\sin^3x-\sin x}{\sin^3x-\sin x} dx\\
&=\int\limits_{0}^{\frac{\pi}{2}}dx\\
&=\frac{\pi}{2}
\end{align*}
I want to know what the name of this rule.

This is actually not a valid rule. To get an intuition why, we can imagine two functions other than sine and cosine that have the same integral over a certain region.
Let the functions be
$$
f_1(x)=\begin{cases}0 & x\leq1\\
1 & x>1\end{cases}
$$
$$
f_2(x)=\begin{cases}1 & x\leq1\\
0 & x>1\end{cases}
$$
Clearly these functions both have an area of 1 when integrated from 0 to 2. But see what happens when we multiply them together:
$$
f_1(x)f_2(x)=\begin{cases}0\cdot1 & x\leq1\\
1\cdot0 & x>1\end{cases}=0
$$
So if we tried to integrate their product, the answer would clearly be zero, showing that we cannot replace one with the other if they are multiplied together. Similar reasoning follows for division.
