Show that $\int_0^\pi f(\sin x)\,\mathrm{d}x = 2\int_0^{\pi/2}f(\sin x) \, \mathrm{d}x$ [closed]

Any tips on how to show that $$\int_0^\pi f(\sin x) \, \mathrm{d}x = 2\int_0^{\pi/2} f(\sin x) \, \mathrm{d}x \text{ ?}$$

Solution 1:

Note that we have

$$\begin{align} \int_0^\pi f(\sin(x))\,dx&=\int_0^{\pi/2} f(\sin(x))\,dx+\int_{\pi/2}^{\pi} f(\sin(x))\,dx\\\\ &=\int_0^{\pi/2} f(\sin(x))\,dx+\underbrace{\int_{\pi/2}^0 f(\sin(\pi-t))\,(-dt)}_{\text{after the substitution}\,\,x=\pi-t}\\\\ &=\int_0^{\pi/2} f(\sin(x))\,dx+\int_0^{\pi/2} f(\sin(t))\,dt\\\\ &=2\int_0^{\pi/2} f(\sin(x))\,dx \end{align}$$

Solution 2:

Split the integral:$$\int_{0}^{\pi}f(\sin(x)) = \int_{0}^{\pi/2}f(\sin(x)) + \int_{\pi/2}^{\pi}f(\sin(x))$$

Shift one integral:

$$\int_{\pi/2}^{\pi}f(\sin(x)) = \int_{0}^{\pi/2}f(\sin(x + \pi/2))$$

Reverse:

$$\int_{0}^{\pi/2}f(\sin(x + \pi/2)) = \int_{0}^{\pi/2}f(\sin((\pi/2 - x) + \pi/2)) = \int_{0}^{\pi/2}f(\sin(\pi - x))$$

But since $\sin(x) = \sin(\pi - x)$ we can simplify:

$$\int_{0}^{\pi}f(\sin(x)) = 2\int_{0}^{\pi/2}f(\sin(x))$$

Solution 3:

$$\sin \left( \frac{\pi}{4} + x \right) = \sin \left( \frac{\pi}{4} - x \right) \text{.} $$

Solution 4:

Generalization:

$$\int_0^{2a}f(x)\ dx=\int_0^af(x)\ dx+\int_a^{2a}f(x)\ dx$$

Now set $2a-x=u$ and what happens if $f(2a-x)=f(x)$