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.
 A: I think you want

$$\int_a^b f(x) dx=\int_a^b f(a+b-x) dx$$

If you input $a=0,b=\pi/2$, using the above property you can "convert" sines to cosines and vice versa due to $\sin x=\cos (\pi/2 -x)$.
But what you have done, as pointed out by others, is not applicable everywhere. If you ''exchange" sines and cosines using the above property, it is totally fine.
A: There is a rule that allows exchanging sines and cosines (and, generally, trig functions and their respective co-functions), but it requires changing them all ... and possibly making other adjustments for non-trig elements.
The "rule" is simply the $u$-substitution $u=\pi/2-x$, which we write thusly:
$$\int_0^{\pi/2}f(x) dx = \int_{\pi/2}^0 f\left(\frac\pi2-u\right)(-du) = \int_0^{\pi/2}f\left(\frac\pi2-u\right)du= \int_0^{\pi/2}f\left(\frac\pi2-x\right)dx \tag{$\star$}$$
where the last step simply replaces the integration variable.
Now, since $\sin x = \cos(\pi/2-x)=\cos u$ and $\cos x = \sin(\pi/2-x) = \sin u$, the effect of $(\star)$ is to "magically" exchange all sines and cosines (and all trig functions and co-functions).
Importantly: Every instance of $\sin x$ must  be  changed to $\cos x$, and vice-versa. You don't get to pick and choose. (Also, any non-trigged instances of $x$ become $\pi/2-x$, which is decidedly non-magical.)
So, use the "rule" with caution.
In particular, the example's replacement of cosines with sines without the vice-versa, is invalid. It is perhaps worth noting that
$$\int_0^{\pi/2}\frac{\sin^3 x - \cos x}{\cos^3 x - \sin x}dx$$ is an improper integral (due to a singularity at $x=0.598\ldots$). WolframAlpha even times-out trying to evaluate it. Evaluating the integral before and after the problem point and adding the results gives a value of about $8.71605$, which is not $\pi/2$.
A: I am not sure such a "rule" exists. If I understand what you have said correctly, then we have
$$\int_0^{\pi/2}\tan xdx=\int_0^{\pi/2}\frac{\sin x}{\cos x}dx
=\int_0^{\pi/2}\frac{\cos x}{\cos x}dx=\pi/2$$
where we used the "rule" in the second equality.
However,
$$\int_0^{\pi/2}\tan xdx=-\ln \cos x\Big|_{x=0}^{x=\pi/2}=\infty.$$
A: This is not a general rule. For example, take
$$\int_0^{\frac{\pi}{2}} \frac{\sin(x)}{x} dx \approx 1.3707621$$
and
$$\int_0^{\frac{\pi}{2}} \frac{\cos(x)}{x} dx$$
which is undefined.
A: This is just plain wrong. Indeed, if you evaluate your original integral numerically, you get a negative answer.
What is correct is this: For any continuous function $f(x,y)$, it is the case that
$$\int_0^{\pi/2} f(\sin\theta,\cos\theta)\,d\theta = \int_0^{\pi/2} f(\cos\theta,\sin\theta)\,d\theta.$$
A: 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.
A: This is not a specific rule. It is the property of definite integral: $\int_a^bf(x)dx=\int_a^bf(a+b-x)dx$ i.e. substitute $x=a+b-x$ everywhere in the integrand as follows
$$\int\limits_{0}^{\frac{\pi}{2}} \frac{\sin^3x-\cos x}{\cos^3x-\sin x} dx$$
$$=\int\limits_{0}^{\frac{\pi}{2}} \frac{\sin^3\left(\frac{\pi}{2}-x\right)-\cos\left(\frac{\pi}{2}-x\right)}{\cos^3\left(\frac{\pi}{2}-x\right)-\sin \left(\frac{\pi}{2}-x\right)} dx$$
$$=\int\limits_{0}^{\frac{\pi}{2}} \frac{\cos^3x-\sin x}{\sin^3x-\cos x} dx$$
Using above property in this case is not useful because it's an improper integral.
