Tính C = tích phân của x^2 cos x dx pi^2 / 4 -2

Câu hỏi:

Tính $C={\int }_0^{\frac{\mathrm{\pi }}{2}}{x}^2\cos xdx$

A. $\frac{{\mathrm{\pi }}^2}{4}-2$

B. $\frac{\mathrm{\pi }}{4}-2$

C. $\frac{{\mathrm{\pi }}^2}{6}+2$

D. $\frac{\mathrm{\pi }}{8}+2$

Trả lời:

Chọn A

Đặt

$\left\{\begin{array}{l}u=x^2\\ dv=\cos xdx\end{array}\Rightarrow \left\{\begin{array}{l}du=2xdx\\ v=\sin x\end{array}\right.\right.$

$$\begin{gathered} {\int }_0^{\frac{\mathrm{\pi }}{2}}{x}^2\cos xdx=x^2\sin x{|}_0^{\frac{\mathrm{\pi }}{2}}-2{\int }_0^{\frac{\mathrm{\pi }}{2}}x\sin xdx \\ =\frac{{\mathrm{\pi }}^2}{4}-2{\int }_0^{\frac{\mathrm{\pi }}{2}}x\sin xdx \end{gathered}$$

* Tính: $I={\int }_0^{\frac{\mathrm{\pi }}{2}}x \sin xdx$

Đặt $\left\{\begin{array}{l}u=x\\ dv=\sin xdx\end{array}\Rightarrow \left\{\begin{array}{l}du=dx\\ v=-\cos x\end{array}\right.\right.$

$$\begin{gathered} I={\int }_0^{\frac{\mathrm{\pi }}{2}}x\sin xdx=-xcos x{|}_0^{\frac{\mathrm{\pi }}{2}}+{\int }_0^{\frac{\mathrm{\pi }}{2}}\cos xdx \\ =-x.\cos x{|}_0^{\frac{\mathrm{\pi }}{2}}+\sin x{|}_0^{\frac{\mathrm{\pi }}{2}}=1 \end{gathered}$$

Thế I = 1 vào C ta được ${\int }_0^{\frac{\mathrm{\pi }}{2}}{x}^2 \cos xdx=\frac{{\mathrm{\pi }}^2}{4}-2$