Tính \(C = \mathop \smallint \nolimits_0^{\frac{{\rm{\pi }}}{2}} {x^2}\cos xdx\)
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Lời giải:
Báo saiĐặt \(\left\{ {\begin{array}{*{20}{l}}
{u = {x^2}}\\
{dv = \cos \;xdx}
\end{array} \Rightarrow \left\{ {\begin{array}{*{20}{l}}
{du = 2xdx}\\
{v = \sin x}
\end{array}} \right.} \right.\)
\(\mathop \smallint \nolimits_0^{\frac{{\rm{\pi }}}{2}} {x^2}\cos \;xdx = {x^2}\sin x|_0^{\frac{{\rm{\pi }}}{2}} - 2\mathop \smallint \nolimits_0^{\frac{{\rm{\pi }}}{2}} x\sin xdx\; = \frac{{{{\rm{\pi }}^2}}}{4} - 2\mathop \smallint \nolimits_0^{\frac{{\rm{\pi }}}{2}} x\sin xdx\)
* Tính: \(I = \mathop \smallint \nolimits_0^{\frac{{\rm{\pi }}}{2}} x\;\sin xdx\)
Đặt \(\left\{ {\begin{array}{*{20}{l}}
{u = x}\\
{dv = \sin \;xdx}
\end{array} \Rightarrow \left\{ {\begin{array}{*{20}{l}}
{du = dx}\\
{v = - \cos \;x}
\end{array}} \right.} \right.\)
\(I = \mathop \smallint \nolimits_0^{\frac{{\rm{\pi }}}{2}} x\sin xdx = - xcos\;x|_0^{\frac{{\rm{\pi }}}{2}} + \mathop \smallint \nolimits_0^{\frac{{\rm{\pi }}}{2}} \cos \;xdx\; = - x.\cos \;x|_0^{\frac{{\rm{\pi }}}{2}} + \sin \;x|_0^{\frac{{\rm{\pi }}}{2}} = 1\)
Thế I = 1 vào C ta được \(\mathop \smallint \nolimits_0^{\frac{{\rm{\pi }}}{2}} {x^2}\;\cos xdx = \frac{{{{\rm{\pi }}^2}}}{4} - 2\)
