a) Xét với \(n > 2\), ta có: \({I_n} = \int\limits_0^{\dfrac{\pi }{2}} {{{\sin }^{n - 1}}x.\sin xdx} \)
Dùng tích phân từng phần với \(u = {\sin ^{n - 1}}x\) và \(dv = \sin xdx\), ta có: \(\left\{ \begin{array}{l}du = \left( {n - 1} \right){\sin ^{n - 2}}x\cos xdx\\v = - \cos x\end{array} \right.\)
\({I_n} = \int\limits_0^{\dfrac{\pi }{2}} {{{\sin }^{n - 1}}x\sin xdx} \)\( = \left. { - \cos x{{\sin }^{n - 1}}x} \right|_0^{\dfrac{\pi }{2}}\) \( + (n - 1)\int\limits_0^{\dfrac{\pi }{2}} {{{\sin }^{n - 2}}x{{\cos }^2}xdx} \)
\( = \left( {n - 1} \right)\int\limits_0^{\dfrac{\pi }{2}} {\left( {{{\sin }^{n - 2}}x - {{\sin }^n}x} \right)dx} \)\( = \left( {n - 1} \right){I_{n - 2}} - \left( {n - 1} \right){I_n}\)
Vậy \({I_n} = \dfrac{{n - 1}}{n}{I_{n - 2}}\)
b) Ta có: \({I_1} = \int\limits_0^{\dfrac{\pi }{2}} {\sin xdx} \)\( = \left. { - \cos x} \right|_0^{\dfrac{\pi }{2}} = 1\).
Suy ra \({I_3} = \dfrac{{3 - 1}}{3}{I_1} = \dfrac{2}{3}.1 = \dfrac{2}{3}\); \({I_5} = \dfrac{{5 - 1}}{5}{I_3} = \dfrac{4}{5}.\dfrac{2}{3} = \dfrac{8}{{15}}\).
Vậy \({I_3} = \dfrac{2}{3},{I_5} = \dfrac{8}{{15}}\).
