Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
\int\limits_0^\pi {x.\cos xdx} \\
\left\{ \begin{array}{l}
u = x\\
v' = \cos x
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
u' = 1\\
v = \sin x
\end{array} \right.\\
\Rightarrow \int\limits_0^\pi {x.\cos xdx} = \mathop {\left. {x.\sin x} \right|}\nolimits_0^\pi - \int\limits_0^\pi {\sin xdx} \\
= \mathop {\left( {x.\sin x + \cos x} \right)}\nolimits_0^\pi \\
= - 2\\
b,\\
\int\limits_0^\pi {x.{{\cos }^2}xdx} = \int\limits_0^\pi {x.\left( {\frac{{\cos 2x + 1}}{2}} \right)} dx\\
= \frac{1}{2}\int\limits_0^\pi {\left( {x + x.\cos 2x} \right)dx} \\
= \frac{1}{2}.\left( {\int\limits_0^\pi {xdx} + \int\limits_0^\pi {x.\cos 2x.dx} } \right)\\
= \frac{1}{2}.\left( {\mathop {\left. {\frac{{{x^2}}}{2}} \right|}\nolimits_0^\pi + \frac{1}{4}\int\limits_0^\pi {2x.\cos 2x.d\left( {2x} \right)} } \right)\\
= \frac{{{\pi ^2}}}{4} + \frac{1}{8}.\int\limits_0^{2\pi } {t.\cos t.dt} \\
= \frac{{{\pi ^2}}}{4} + \frac{1}{8}.0\\
= \frac{{{\pi ^2}}}{4}
\end{array}\)
