Cho hàm số \(f\left( x \right)\) có \(f\left( {\dfrac{\pi }{2}} \right) = 2\) và \(f'\left( x \right) = x\sin x\). Giả sử rằng \(\int\limits_0^{\frac{\pi }{2}} {\cos x.f\left( x \right)dx}  = \dfrac{a}{b} - \dfrac{{{\pi ^2}}}{c}\) (với \(a,\,\,b,\,\,c\) là các số nguyên dương, \(\dfrac{a}{b}\) tối giản). Khi đó \(a + b + c\) bằng:

Đặt \(\left\{ \begin{array}{l}u = f\left( x \right)\\dv = \cos xdx\end{array} \right. \Rightarrow \left\{ \begin{array}{l}du = f'\left( x \right)dx = x\sin xdx\\v = \sin x\end{array} \right.\).

Khi đó ta có:

\(\begin{array}{l}\int\limits_0^{\frac{\pi }{2}} {\cos x.f\left( x \right)dx} \\ = \left. {\sin x.f\left( x \right)} \right|_0^{\frac{\pi }{2}} - \int\limits_0^{\frac{\pi }{2}} {x{{\sin }^2}xdx} \\ = \sin \dfrac{\pi }{2}.f\left( {\dfrac{\pi }{2}} \right) - \int\limits_0^{\frac{\pi }{2}} {x\dfrac{{1 - \cos 2x}}{2}dx} \\ = 2 - \dfrac{1}{2}\left( {\int\limits_0^{\frac{\pi }{2}} {xdx}  - \int\limits_0^{\frac{\pi }{2}} {x\cos 2xdx} } \right)\\ = 2 - \dfrac{1}{2}\left( {\left. {\dfrac{{{x^2}}}{2}} \right|_0^{\frac{\pi }{2}} - I} \right)\\ = 2 - \dfrac{1}{2}\left( {\dfrac{{{\pi ^2}}}{8} - I} \right)\\ = 2 - \dfrac{{{\pi ^2}}}{{16}} + \dfrac{I}{2}\end{array}\)

Xét tích phân \(I = \int\limits_0^{\frac{\pi }{2}} {x\cos 2xdx} \).

Đặt \(\left\{ \begin{array}{l}u = x\\dv = \cos 2xdx\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}du = dx\\v = \dfrac{{\sin 2x}}{2}\end{array} \right.\), khi đó ta có:

\(\begin{array}{l}I = \left. {x.\dfrac{{\sin 2x}}{2}} \right|_0^{\frac{\pi }{2}} - \dfrac{1}{2}\int\limits_0^{\frac{\pi }{2}} {\sin 2xdx} \\I = \dfrac{\pi }{2}.\dfrac{{\sin \pi }}{2} - 0 + \dfrac{1}{2}.\left. {\dfrac{{\cos 2x}}{2}} \right|_0^{\frac{\pi }{2}}\\I = \dfrac{1}{4}\left( {\cos \pi  - \cos 0} \right)\\I = \dfrac{1}{4}\left( { - 1 - 1} \right) =  - \dfrac{1}{2}\end{array}\)

Do đó \(\int\limits_0^{\frac{\pi }{2}} {\cos x.f\left( x \right)dx}  = 2 - \dfrac{{{\pi ^2}}}{{16}} - \dfrac{1}{4} = \dfrac{7}{4} - \dfrac{{{\pi ^2}}}{{16}}\)

\( \Rightarrow a = 7,\,\,b = 4,\,\,c = 16\)

Vậy \(a + b + c = 7 + 4 + 16 = 27\).