Problem
Let $f(x)$ be a continuous function satisfying that $f(x)-\cos^2 x=\dfrac{1}{\pi}\displaystyle\int_0^{\frac{\pi}{4}}f(2x){\rm d}x$. Evaluate $\displaystyle\int_0^\frac{\pi}{2}f(x){\rm d}x$.
Solution
Denote $\displaystyle \int_0^\frac{\pi}{2}f(x){\rm d}x=:I.$ Make a substitution. Let $x=:\dfrac{u}{2}$. Then ${\rm d}x=\dfrac{1}{2}{\rm d}u$. It follows that $$\int_0^{\frac{\pi}{4}}f(2x){\rm d}x=\frac{1}{2}\int_0^{\frac{\pi}{2}}f(u){\rm d}u=\frac{I}{2}.$$ Thus, according to the given equality, we have$$f(x)-\cos^2 x=\frac{I}{2\pi}$$ Integrate the both sides from $0$ to $\dfrac{\pi}{2}$. We obtain $$\int_0^\frac{\pi}{2}f(x){\rm d}x-\int_0^\frac{\pi}{2}\cos^2 x{\rm d}x=\int_0^\frac{\pi}{2}\frac{I}{2\pi}{\rm d}x,$$which implies $$I-\frac{\pi}{4}=\frac{I}{2\pi}\cdot \left(\frac{\pi}{2}-0\right).$$ As a result, $$I=\frac{\pi}{3}.$$