Đáp án:
B=$\frac{-2}{3}$
Giải thích các bước giải:
\(
\begin{array}{l}
B = \int_{\rm{0}}^\pi {{\mathop{\rm s}\nolimits} {\rm{inx}}{\rm{.cos2xdx}}} \\
Đặt :\left\{ {\begin{array}{*{20}c}
{\cos 2x = u} \\
{\sin xdx = dv} \\
\end{array}} \right. \Leftrightarrow \left\{ {\begin{array}{*{20}c}
{du = - 2\sin 2xdx} \\
{v = - \cos x} \\
\end{array}} \right. \\
\Rightarrow B = \left. { - \cos x.\cos 2x} \right|_0^\pi - \int_0^\pi { - \cos x.( - 2\sin 2xdx)} \\
= \left. { - \cos x.\cos 2x} \right|_0^\pi - 2\int_0^\pi {\cos x.\sin 2xdx} \\
Đặt:\left\{ {\begin{array}{*{20}c}
{\sin 2x = u} \\
{\cos xdx = dv} \\
\end{array}} \right. \Leftrightarrow \left\{ {\begin{array}{*{20}c}
{du = 2\cos 2xdx} \\
{v = \sin x} \\
\end{array}} \right. \\
\Rightarrow B = \left. { - \cos x.\cos 2x} \right|_0^\pi - 2(\left. {\sin x.\sin 2x} \right|_0^\pi - \int_{\rm{0}}^\pi {{\mathop{\rm s}\nolimits} {\rm{inx}}{\rm{.2cos2xdx}}} ) \\
= \left. { - \cos x.\cos 2x} \right|_0^\pi - 2\left. {\sin x.\sin 2x} \right|_0^\pi + 4\int_{\rm{0}}^\pi {{\mathop{\rm s}\nolimits} {\rm{inx}}{\rm{.2cos2xdx}}} \\
= \left. { - \cos x.\cos 2x} \right|_0^\pi - 2\left. {\sin x.\sin 2x} \right|_0^\pi + 4B \\
= > 3B = \left. {\cos x.\cos 2x} \right|_0^\pi + 2\left. {\sin x.\sin 2x} \right|_0^\pi \\
= \cos \pi .\cos 2\pi + 2\sin \pi .\sin 2\pi - \cos 0.\cos 2.0 - \sin 0.\sin 2.0 \\
= - 2 \\
= > B = \frac{{ - 2}}{3} \\
\end{array}
\)
