Cho hàm số A I=32/2 B I=31 C I=71/6 D I=32

+ Tính $\underset{0}{\overset{\frac{\pi }{2}}{\int }}f\left(\sin x\right)\cos xdx$. Đặt $\sin x=t\Rightarrow \cos xdx=dt$. Đổi cận $\left\{\begin{array}{l}x=0\Rightarrow t=0\\ x=\frac{\pi }{2}\Rightarrow t=1\end{array}\right.$

Do đó $\underset{0}{\overset{\frac{\pi }{2}}{\int }}f\left(\sin x\right)\cos xdx=\underset{0}{\overset{1}{\int }}f\left(t\right)dt=\underset{0}{\overset{1}{\int }}\left(5-t\right)dt=\left(5t-\frac{t^2}{2}\right)\left|\begin{array}{l}{}^1\\ {}_0\end{array}\right.=\frac{9}{2}$

+ Tính $\underset{0}{\overset{1}{\int }}f\left(3-2x\right)dx$. Đặt $t=3-2x\Rightarrow dt=-2dx\Rightarrow dx=\frac{-dt}{2}$

Đổi cận $\left\{\begin{array}{l}x=0\Rightarrow t=3\\ x=1\Rightarrow t=1\end{array}\right.$

Do đó $\underset{0}{\overset{1}{\int }}f\left(3-2x\right)dx=\underset{3}{\overset{1}{\int }}f\left(t\right).\frac{-dt}{2}=\frac{1}{2}\underset{1}{\overset{3}{\int }}f\left(t\right)dt=\frac{1}{2}\underset{1}{\overset{3}{\int }}\left(x^2+3\right)dt=\frac{1}{2}\left(\frac{x^3}{3}+3x\right)\left|\begin{array}{l}{}^3\\ {}_1\end{array}\right.=\frac{22}{3}$

Vậy $I=2.\frac{9}{2}+3.\frac{22}{3}=31$

Chọn đáp án B