Question

[minh trinh]

Tính tích phân I=\(\int\limits^{\pi}_0\)\(x^2cos2xdx\) bằng cách đặt \(\left\{{}\begin{matrix}u=x^2\\dv=cos2xdx\end{matrix}\right.\).Mệnh đề nào dưới đây đúng?

A. \(I=\dfrac{1}{2}x^2sin2x|^{^{\pi}_0}-\int\limits^{\pi}_0xsin2xdx\)

B. \(I=\dfrac{1}{2}x^2sin2x|^{^{\pi}_0}-2\int\limits^{\pi}_0xsin2xdx\)

C. \(I=\dfrac{1}{2}x^2sin2x|^{^{\pi}_0}+\int\limits^{\pi}_0xsin2xdx\)

D. \(I=\dfrac{1}{2}x^2sin2x|^{^{\pi}_0}+2\int\limits^{\pi}_0xsin2xdx\)

Answer 1

[Nguyễn Việt Lâm]

\(\left\{{}\begin{matrix}u=x^2\\dv=cos2xdx\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}du=2xdx\\v=\dfrac{1}{2}sin2x\end{matrix}\right.\)

\(\Rightarrow I=\dfrac{1}{2}x^2sin2x|^{\pi}_0-\int\limits^{\pi}_0x.sin2xdx\)