Bài tập 3.23 trang 172 SBT Toán 12

a) Xét với n > 2, ta có:

\({I_n} = \mathop \smallint \limits_0^{\frac{\pi }{2}} {\sin ^{n - 1}}x\sin xdx,n \in {N^ * }.\)

Đặt \(\left\{ \begin{array}{l}
u = {\sin ^{n - 1}}x\\
dv = \sin xdx
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
du = (n - 1){\sin ^{n - 2}}x.\cos xdv\\
v =  - \cos x
\end{array} \right.\)

Khi đó:

\(\begin{array}{l}
{I_n} = \left. { - {{\sin }^{n - 1}}x.\cos x} \right|_0^{\frac{\pi }{2}} + \left( {n - 1} \right)\int\limits_0^{\frac{\pi }{2}} {{{\sin }^{n - 2}}x{{\cos }^2}xdx} \\
 = \left( {n - 1} \right)\int\limits_0^{\frac{\pi }{2}} {{{\sin }^{n - 2}}x\left( {1 - {{\sin }^2}x} \right)dx} \\
 = \left( {n - 1} \right)\left( {{I_{n - 2}} - {I_n}} \right)\\
 \Rightarrow {I_n} = \left( {n - 1} \right){I_{n - 2}} - \left( {n - 1} \right){I_n}\\
 \Rightarrow {I_n} = \frac{{n - 1}}{n}{I_{n - 2}}
\end{array}\)

b) 

\(\begin{array}{l}
{I_3} = \frac{2}{3}{I_1} = \frac{2}{3}\int\limits_0^{\frac{\pi }{2}} {\sin xdx =  - \frac{2}{3}\left. {\cos x} \right|_0^{\frac{\pi }{2}}}  = \frac{2}{3}\\
{I_5} = \frac{4}{5}{I_3} = \frac{4}{5}.\frac{2}{3} = \frac{8}{{15}}
\end{array}\)

-- Mod Toán 12 HỌC247