Tính tích phân sau: \(\mathop \smallint \nolimits_0^{\frac{{\rm{\pi }}}{2}} \sin \;3x.\cos \;xdx\)
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Lời giải:
Báo sai\(\begin{array}{l}
\mathop \smallint \nolimits_0^{\frac{{\rm{\pi }}}{2}} \sin 3x\;\cos xdx\\
= \frac{1}{2}\mathop \smallint \nolimits_0^{\frac{{\rm{\pi }}}{2}} \left[ {\sin 4x + \sin 2x} \right]dx\\
= \frac{1}{2}\left[ {\mathop \smallint \nolimits_0^{\frac{{\rm{\pi }}}{2}} \sin 4xdx + \mathop \smallint \nolimits_0^{\frac{{\rm{\pi }}}{2}} \sin 2xdx} \right]\\
= \frac{1}{2}\left[ { - \frac{1}{4}\cos 4x - \frac{1}{2}\cos 2x} \right]\left| {\begin{array}{*{20}{c}}
{\frac{{\rm{\pi }}}{2}}\\
0
\end{array}} \right.\\
= \frac{1}{2}\left[ {\left( { - \frac{1}{4}\cos 2{\rm{\pi }} - \frac{1}{2}cos\pi } \right) - \left( { - \frac{1}{4}\cos 0 - \frac{1}{2}\cos 0} \right)} \right]\\
= \frac{1}{2}\left( { - \frac{1}{4} + \frac{1}{2} + \frac{1}{4} + \frac{1}{2}} \right) = \frac{1}{2}
\end{array}\)