Cho hàm số $y=f\left( x \right)=\left\{ \begin{aligned} &...

+ Tính $\int\limits_{0}^{\dfrac{\pi }{2}}{f\left( \sin x \right)\cos xdx}$. Đặt $\sin x=t\Rightarrow \cos xdx=dt$. Đổi cận $\left\{ \begin{aligned}
& x=0\Rightarrow t=0 \\
& x=\dfrac{\pi }{2}\Rightarrow t=1 \\
\end{aligned} \right.$
Do đó $\int\limits_{0}^{\dfrac{\pi }{2}}{f\left( \sin x \right)\cos xdx}=\int\limits_{0}^{1}{f\left( t \right)dt}=\int\limits_{0}^{1}{\left( 5-t \right)dt}=\left( 5t-\dfrac{{{t}^{2}}}{2} \right)\left| \begin{aligned}
& ^{1} \\
& _{0} \\
\end{aligned} \right.=\dfrac{9}{2}$
+ Tính $\int\limits_{0}^{1}{f\left( 3-2x \right)dx}$. Đặt $t=3-2x\Rightarrow dt=-2dx\Rightarrow dx=\dfrac{-dt}{2}$
Đổi cận $\left\{ \begin{aligned}
& x=0\Rightarrow t=3 \\
& x=1\Rightarrow t=1 \\
\end{aligned} \right.$
Do đó $\int\limits_{0}^{1}{f\left( 3-2x \right)dx}=\int\limits_{3}^{1}{f\left( t \right).\dfrac{-dt}{2}}=\dfrac{1}{2}\int\limits_{1}^{3}{f\left( t \right)dt}=\dfrac{1}{2}\int\limits_{1}^{3}{\left( {{x}^{2}}+3 \right)dt}=\dfrac{1}{2}\left( \dfrac{{{x}^{3}}}{3}+3x \right)\left| \begin{aligned}
& ^{3} \\
& _{1} \\
\end{aligned} \right.=\dfrac{22}{3}$
Vậy $I=2.\dfrac{9}{2}+3.\dfrac{22}{3}=31$