Đáp án:
$D$
Giải thích các bước giải:
$\sin xf(\cos x)+\cos x f(\sin x)=\sin 2x -\dfrac{1}{2} \sin^32x\\ \displaystyle\int\limits^\tfrac{\pi}{4}_0 \sin xf(\cos x) \, dx+\displaystyle\int\limits^\tfrac{\pi}{4}_0 \cos x f(\sin x) \, dx=\displaystyle\int\limits^\tfrac{\pi}{4}_0 \sin 2x \, dx -\displaystyle\int\limits^\tfrac{\pi}{4}_0 \dfrac{1}{2} \sin^32x \, dx\\ \Leftrightarrow -\displaystyle\int\limits^\tfrac{\pi}{4}_0 f(\cos x) \, d(\cos x)+\displaystyle\int\limits^\tfrac{\pi}{4}_0 f(\sin x) \, d(\sin x)=2\displaystyle\int\limits^\tfrac{\pi}{4}_0 \sin x \cos x\, dx -\dfrac{1}{2} .8\displaystyle\int\limits^\tfrac{\pi}{4}_0 \sin^3x\cos^3x \, dx\\ \Leftrightarrow -\displaystyle\int\limits^\tfrac{\sqrt{2}}{2}_1 f(x) \, dx+\displaystyle\int\limits^\tfrac{\sqrt{2}}{2}_0 f(x) \, dx=2\displaystyle\int\limits^\tfrac{\pi}{4}_0 \sin x \, d(\sin x) -4\displaystyle\int\limits^\tfrac{\pi}{4}_0 \sin^3x(1-\sin^2x) \, d(\sin x)\\ \Leftrightarrow \displaystyle\int\limits^1_ \tfrac{\sqrt{2}}{2} f(x) \, dx+\displaystyle\int\limits^\tfrac{\sqrt{2}}{2}_0 f(x) \, dx=2\displaystyle\int\limits^\tfrac{\sqrt{2}}{2}_0 x \, dx +4\displaystyle\int\limits^0_\tfrac{\sqrt{2}}{2} x^3(1-x^2) \, dx\\ \Leftrightarrow \displaystyle\int\limits^1_0 f(x) \, dx=x^2\Bigg\vert^\tfrac{\sqrt{2}}{2}_0 -4\left(\dfrac{x^4}{4}-\dfrac{x^6}{6}\right)\Bigg\vert^0_\tfrac{\sqrt{2}}{2} \\ \Leftrightarrow \displaystyle\int\limits^1_0 f(x) \, dx=\dfrac{1}{3}\\ \Rightarrow D$
