Tính tích phân sau: I=∫x(sin2x+2x) dx

Question

Tính tích phân sau:

\(I=\int _{0}^{1}x\left(\sin 2x+2x\right){\rm d}x \)

Answer 1

\(I=\int _{0}^{1}x\left(\sin 2x+2x\right)dx  \)
\(I=\int _{0}^{1}x\left(\sin 2x+2x\right)dx =\int _{0}^{1}x\sin 2xdx +2\int _{0}^{1}x^{2} dx\)  

\( I=\int _{0}^{1}x\sin 2xdx  \)

\( Đặt \left\{\begin{array}{l} {u=x} \\ {dv=\sin 2xdx} \end{array}\right. \Rightarrow \left\{\begin{array}{l} {du=dx} \\ {v=-\frac{1}{2} \cos 2x} \end{array}\right. \)
\(I=-\frac{1}{2} x\cos 2x\left|\begin{array}{l} {1} \\ {0} \end{array}\right. +\frac{1}{2} \int _{0}^{1}\cos 2xdx  \)
\(=-\frac{1}{2} \cos 2+\frac{1}{4} \sin 2x\left|\begin{array}{l} {1} \\ {0} \end{array}\right. =\frac{1}{4} \sin 2-\frac{1}{2} \cos 2 \)
Vậy: \(\int _{0}^{1}x\left(\sin 2x+2x\right)dx =\frac{1}{4} \sin 2-\frac{1}{2} \cos 2+2\int _{0}^{1}x^{2} dx \)

\(=\frac{1}{4} \sin 2-\frac{1}{2} \cos 2+\frac{2}{3} \)