Prove that for any continuous on $[0,1]$ function: $\int_0^{\pi\over2}f(\sin x)\,dx = \int_0^{\pi\over2}f(\cos x)\,dx$

Question

Prove that for any continuous on $[0,1]$ the following equality holds: $$ \int_0^{\pi\over2}f(\sin x)\,dx = \int_0^{\pi\over2}f(\cos x)\,dx $$

I've basically used the King's rule, namely: $$ \int_a^b f(x)\,dx= \int_a^bf(a+b - x)\,dx $$ So: $$ \int_0^{\pi\over2}f(\sin x)\,dx = \int_0^{\pi\over2}f\left(\sin \left({\pi\over 2} - x\right)\right)\,dx $$ But: $$ \sin \left({\pi\over 2} - x\right) = \cos(x) $$ Thus: $$ \int_0^{\pi\over2}f\left(\sin \left({\pi\over 2} - x\right)\right)\,dx = \int_0^{\pi\over2}f(\cos x)\,dx $$

I get why we need $f(x)$ to be continuous, however, I don't really understand why we require $f(x)$ to be continuous specifically on the range $[0, 1]$. Could you please explain that?

Answer 1

If $0\le x\le \frac \pi 2$ then $0\le\cos x\le 1$ and $0\le \sin x \le 1.$

Answer 2

We know that continuous functions are integrable. We can formulate a weaker version, assuming that the integrals exist, but with continuity we have this guaranteed. And arguments of $f$ are values of $\sin$ and $\cos$ in $[0,\pi/2]$ and this in the interval $[0,1]$.

Answer 3

Both integrals are equal to $\int_0^1 \frac{f(u)}{\sqrt{1-u^2}} du$.

In the first integral, use the substitution $u=\sin x$, and in the second $u=\cos x$.