Lời giải:Ta có: \(I = \int\limits_0^{\dfrac{\pi }{2}} {x.\cos \left( {a - x} \right)\,dx} \)\(\,= - \int\limits_0^{\dfrac{\pi }{2}} {x\,d\left( {\sin \left( {a - x} \right)} \right)} \)
Đặt \(\left\{ \begin{array}{l}u = x\\dv = d\left( {\sin \left( {a - x} \right)} \right)\end{array} \right. \)
\(\Rightarrow \left\{ \begin{array}{l}du = dx\\v = \sin \left( {a - x} \right)\end{array} \right.\)
Khi đó ta có:
\(I = - \left( {x\sin \left( {a - x} \right)} \right)\left| {_{_{\scriptstyle\atop{\scriptstyle\atop\scriptstyle0}}}^{\dfrac{\pi }{2}}} \right. + \int\limits_0^{\dfrac{\pi }{2}} {\sin \left( {a - x} \right)} \,dx\)
\(= - \dfrac{\pi }{2}\sin \left( {a - \dfrac{\pi }{2}} \right) + \int\limits_0^{\dfrac{\pi }{2}} d \left( {\cos \left( {a - x} \right)} \right)\)
\( = - \dfrac{\pi }{2}\sin \left( {a - \dfrac{\pi }{2}} \right) + \cos \left( {a - x} \right)\left| \begin{array}{l}^{\dfrac{\pi }{2}}\\_0\end{array} \right. \)
\(= - \dfrac{\pi }{2}\sin \left( {a - \dfrac{\pi }{2}} \right) + \cos \left( {a - \dfrac{\pi }{2}} \right) - \cos a\)
\( = \dfrac{\pi }{2}\cos a + \sin a - \cos a \)
\(= \left( {\dfrac{\pi }{2} - 1} \right)\,\cos a + \sin a\)
