If f:RR is a continuous function satisfying _0^/2f(2x) x,dx+_0^/4f(2x) x,dx=0, then the va…

Question

If $f:\mathbb{R}	o\mathbb{R}$ is a continuous function satisfying $\displaystyle\int_0^{\pi/2}f(\sin2x)\sin x\,dx+\alpha\int_0^{\pi/4}f(\cos2x)\cos x\,dx=0$, then the value of $\alpha$ is:

Accepted Answer

Split the first integral at $\pi/4$: $\displaystyle\int_0^{\pi/2}f(\sin2x)\sin x\,dx = \int_0^{\pi/4}f(\sin2x)\sin x\,dx+\int_{\pi/4}^{\pi/2}f(\sin2x)\sin x\,dx$.

In the second piece, substitute $x=t+\pi/4$ so it becomes $\displaystyle\int_0^{\pi/4}f(\cos2t)\sin\left(t+\frac\pi4\right)dt$ (using $\sin(2x)=\sin(2t+\pi/2)=\cos2t$).

Combining all terms over $[0,\pi/4]$ with integrand $f(\cos2x)$:

$$\int_0^{\pi/4}f(\cos2x)\left[\sin\left(\frac\pi4-x\right)+\sin\left(x+\frac\pi4\right)+\alpha\cos x\right]dx = 0$$

Since $\sin(\pi/4-x)+\sin(x+\pi/4) = \sqrt2\cos x$, this requires $\sqrt2+\alpha=0$, so $\alpha=-\sqrt2$.

Solutions · 1