Đáp án:
\[I = \frac{3}{5}\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\sin x.\sin y = \frac{1}{2}\left( {\sin \left( {x + y} \right) + \sin \left( {x - y} \right)} \right)\\
I = \int\limits_0^{\frac{\pi }{2}} {\sin 3x.\sin 2xdx} = \int\limits_0^{\frac{\pi }{2}} {\frac{1}{2}\left( {\sin \left( {3x + 2x} \right) + \sin \left( {3x - 2x} \right)} \right)dx} \\
= \frac{1}{2}\int\limits_0^{\frac{\pi }{2}} {\left( {\sin 5x + \sin x} \right)dx} = \mathop {\left. {\frac{1}{2}\left( { - \frac{{\cos 5x}}{5} - \cos x} \right)} \right|}\nolimits_0^{\frac{\pi }{2}} \\
= - \frac{1}{2}\left( {\frac{{\cos \frac{{5\pi }}{2}}}{5} + \cos \frac{\pi }{2} - \frac{{\cos 0}}{5} - \cos 0} \right)\\
= \frac{3}{5}
\end{array}\)
