Đáp án:
\(
I = \frac{{x^2 \sin 3x}}{3} + \frac{{2x.\cos 3x}}{9} + \frac{{2\sin 3x}}{{27}}
\)
Giải thích các bước giải:
\(
\begin{array}{l}
I = \int {x^2 .\cos 3xdx} \\
Đặt :\left\{ {\begin{array}{*{20}c}
{x^2 = u} \\
{\cos 3xdx = dv} \\
\end{array}} \right. \Leftrightarrow \left\{ {\begin{array}{*{20}c}
{du = 2xdx} \\
{v = \frac{{\sin 3x}}{3}} \\
\end{array}} \right. \\
= > I = \frac{{x^2 .\sin 3x}}{3} - \int {\frac{{\sin 3x}}{3}.2xdx} \\
= \frac{{x^2 \sin 3x}}{3} - \frac{2}{3}\int {x.\sin 3xdx} \\
Đặt:\left\{ {\begin{array}{*{20}c}
{x = u} \\
{\sin 3xdx = dv} \\
\end{array}} \right. \Leftrightarrow \left\{ {\begin{array}{*{20}c}
{du = dx} \\
{v = \frac{{ - \cos 3x}}{3}} \\
\end{array}} \right. \\
= > I = \frac{{x^2 \sin 3x}}{3} - \frac{2}{3}(\frac{{ - x.\cos 3x}}{3} - \int {\frac{{ - \cos 3x}}{3}dx)} \\
= \frac{{x^2 \sin 3x}}{3} + \frac{{2x.\cos 3x}}{9} + \frac{2}{9}\int {\cos 3xdx} \\
= \frac{{x^2 \sin 3x}}{3} + \frac{{2x.\cos 3x}}{9} + \frac{2}{9}.\frac{{\sin 3x}}{3} \\
= \frac{{x^2 \sin 3x}}{3} + \frac{{2x.\cos 3x}}{9} + \frac{{2\sin 3x}}{{27}} \\
\end{array}
\)
