Show that if $f$ is continuous on [0,1], then: $\int_0^\frac\pi 2 f(\sin x)dx=\int_0^\frac\pi 2 f(\cos x)dx= \frac12\int_0^\pi f(\sin x)dx$

Also part of the problem:

Show that: $$\int_0^{n\pi} f(\cos^2 x)dx=n\int_0^{\pi} f(\cos^2 x)dx$$ It may be important to note that I have taken a few semester of calculus prior to this, and that this problem was brought up in my math analysis course. Additionally, my class has only taught as far as Riemannian integrals.

My work: A) For the first part of this problem ($\int_0^\frac\pi 2 f(\sin x)dx=\int_0^\frac\pi 2 f(\cos x)= \frac12\int_0^\pi f(\sin x)$ ), I intuitively know that the areas under the curve will be equal to each other because the values at the endpoints will be the same, provided that $f(\sin x)=\sin x$ and $f(\cos x)=\cos x$. However, I have two dilemmas:

1) The fact that the situation deals with a function of $\sin x$, not $\sin x$ itself. To give an example, $f(a)$ could be equal to $a^2 + 8a^{-1} + \csc a^3$ and I feel like this completely flaws my logic (note that the above function was just an example to show how arbitrary the function is). In the grand scheme of things, I don't believe this will prove to be too much of an issue, I just don't know how I can prove that it won't be an issue

2) I am unaware of how to definitively prove that the area under each curve will be the same. Suppose $f(a)= a$ for the sake of argument, then the value $f(x)$ at each of the endpoints will be the same, but this does not tell anything about the areas under the curve

B) I believe that I understand the second part of the problem. Because $x^2$ will never be negative, the graph of $f(x)$ will either always be negative or always be positive. Therefore, the value of the integral is entirely dependent on $n$.

I am primarily concerned with the two dilemmas in the first part of the problem, but any additional assistance is always welcome!

Solution 1:

Use the fact, $\int_a^{b}f(x)=\int_a^bf(a+b-x)$

$\int_0^\frac\pi 2 f(\sin x)dx=\int_0^\frac\pi 2 f(\sin(\frac{\pi}{2}+0-x))dx= \int_0^\frac{\pi}{2} f(\cos)dx$

Use the fact $\int_c^{a+b}f(x) dx=\int_a^cf(x)dx +\int_c^b f(x)dx$

$\int_0^\pi f(\sin x)dx=\int_0^{\frac\pi 2+\frac\pi 2} f(\sin x)dx=\int_0^{\frac\pi 2} f(\sin x)dx+\int_0^{\frac\pi 2} f(\sin x)dx=2\int_0^{\frac\pi 2} f(\sin x)dx$

Therefore, $\int_0^{\frac\pi 2} f(\sin x)dx=\frac{1}{2}\int_0^{\pi} f(\sin x)dx$

Edit:

Proof of $\int_a^{b}f(x)=\int_a^bf(a+b-x)$

Proof: Substitute $y=a+b-x$ then $dy=-dx$

when $x\rightarrow a$ then $y\rightarrow b$ and when $x\rightarrow b$, then $y\rightarrow a$

Therefore,

$\int_a^bf(a+b-x)=\int_b^{a}f(y)(-dy)=-\int_b^{a}f(y)dy=\int_a^{b}f(y)dy=\int_a^{b}f(x)(dx)=$

Solution 2:

\begin{align*}\int_0^{n\pi}f(\cos^2(x))\mathrm{d}x &= \sum_{i=0}^{n-1}\int_{i\pi}^{(i+1)\pi}f(\cos^2(x))\mathrm{d}x \\ &= \sum_{i=0}^{n-1}\int_0^\pi f(\cos^2(x))\mathrm{d}x \\ &=n\int_0^\pi f(\cos^2(x))\mathrm{d}x \end{align*}

You should ask yourself why the third equality is true.