Biết π 2 ∫ 0 sin 5 x sin 2 x d x = a b với a , b ∈ Z , a < b . Tính 1 a/ a + b .

Chọn A

\[\int\limits_0^{\frac{\pi }{2}} {\sin 5x\sin 2xdx}  =  - \frac{1}{2}\int\limits_0^{\frac{\pi }{2}} {\left( {\cos 7x - \cos 3x} \right)dx} \]\[ = \left. { - \frac{1}{2}\left( {\frac{{\sin 7x}}{7} - \frac{{\sin 3x}}{3}} \right)} \right|_0^{\frac{\pi }{2}} =  - \frac{2}{{21}}\].

Suy ra \(a =  - 2;b = 21\). Do đó \(\frac{1}{2}a + b = 20\).