$\int\limits_0^{\pi / 2} \sin ^7 x d x=$

$\frac{37}{184}$ $\frac{17}{45}$ $\frac{16}{35}$ $\frac{16}{45}$

$\frac{16}{35}$

$I=\frac{6 . 4 . 2}{7 . 5 . 3 . 1}=\frac{16}{35}$ Hence (3) is the correct answer.