$3)2f(x)+f\left(\dfrac{\pi}{2}-x\right)=3\sin x\\ \Rightarrow \displaystyle\int\limits^{\tfrac{\pi}{3}}_{\tfrac{\pi}{6}} 2f(x) \, dx+\displaystyle\int\limits^{\tfrac{\pi}{3}}_{\tfrac{\pi}{6}} f\left(\dfrac{\pi}{2}-x\right) \, dx=\displaystyle\int\limits^{\tfrac{\pi}{3}}_{\tfrac{\pi}{6}} 3\sin x \, dx\\ \Leftrightarrow \displaystyle\int\limits^{\tfrac{\pi}{3}}_{\tfrac{\pi}{6}} 2f(x) \, dx+I=\dfrac{3\sqrt{3}-3}{2}\\ I=\displaystyle\int\limits^{\tfrac{\pi}{3}}_{\tfrac{\pi}{6}} f\left(\dfrac{\pi}{2}-x\right) \, dx\\ u=\dfrac{\pi}{2}-x \Rightarrow du=-dx\\ \text{Đổi cận}\\ \begin{array}{|c|ccccccc|} \hline x&\dfrac{\pi}{6}&&\dfrac{\pi}{3}\\\hline u&\dfrac{\pi}\\{3}&&\dfrac{\pi}{6}\\\hline\end{array}\\ I=-\displaystyle\int\limits^{\tfrac{\pi}{6}}_{\tfrac{\pi}{3}} f(u) \, du\\ =\displaystyle\int\limits^{\tfrac{\pi}{3}}_{\tfrac{\pi}{6}} f(u) \, du\\ =\displaystyle\int\limits^{\tfrac{\pi}{3}}_{\tfrac{\pi}{6}} f(x) \, dx\\ \Rightarrow 2\displaystyle\int\limits^{\tfrac{\pi}{3}}_{\tfrac{\pi}{6}} f(x) \, dx+\displaystyle\int\limits^{\tfrac{\pi}{3}}_{\tfrac{\pi}{6}} f(x) \, dx=\dfrac{3\sqrt{3}-3}{2}\\ \Leftrightarrow \displaystyle\int\limits^{\tfrac{\pi}{3}}_{\tfrac{\pi}{6}} f(x) \, dx=\dfrac{\sqrt{3}-1}{2}\\ 4)I=\displaystyle\int\limits^{2}_{1} (2x+1)f'(x) \, dx\\ u=2x+1 \Rightarrow du=2dx\\ dv=f'(x) \, dx \Rightarrow v=f(x)\\ I=(2x+1)f(x)\Bigg\vert^{2}_{1}-2\displaystyle\int\limits^{2}_{1} f(x) \, dx\\ =5f(2)-3f(1)-2\displaystyle\int\limits^{2}_{1} f(x) \, dx\\ f(1)=4;f(x)+f(3-x)=9\\ \Rightarrow f(1)+f(3-1)=9\\ \Rightarrow f(2)=5\\ f(x)+f(3-x)=9\\ \Rightarrow \displaystyle\int\limits^{2}_{1} f(x) \, dx + \displaystyle\int\limits^{2}_{1} f(3-x) \, dx=\displaystyle\int\limits^{2}_{1} 9 \, dx\\ \Leftrightarrow \displaystyle\int\limits^{2}_{1} f(x) \, dx + \displaystyle\int\limits^{2}_{1} f(x) \, dx=9\\ \Leftrightarrow \displaystyle\int\limits^{2}_{1} f(x) \, dx =\dfrac{9}{2}\\ \Rightarrow I=4$