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

Đặt \(\left\{ {\begin{array}{*{20}{c}}{u = f(x)}\\{dv = cosxdx}\end{array}} \right. \Rightarrow \left\{ {\begin{array}{*{20}{c}}{du = f\prime (x)dx = xsinxdx}\\{v = sinx}\end{array}} \right.\)

Khi đó ta có:

\(\int\limits_0^{\frac{\pi }{2}} {cosx.f(x)dx = sinx.f(x)\left| {_0^{\frac{\pi }{2}}} \right.} - \int\limits_0^{\frac{\pi }{2}} {xsi{n^2}xdx} \)

\[ = sin\frac{\pi }{2}.f\left( {\frac{\pi }{2}} \right) - \int\limits_0^{\frac{\pi }{2}} {x\frac{{1 - cos2x}}{2}} dx\]

\[ = 2 - \frac{1}{2}\left( {\int\limits_0^{\frac{\pi }{2}} {xdx - \int\limits_0^{\frac{\pi }{2}} {xcos2xdx} } } \right)\]

\[\begin{array}{l} = 2 - \frac{1}{2}\left( {\frac{{{x^2}}}{2}\left| {_0^{\frac{\pi }{2}} - I} \right.} \right)\\ = 2 - \frac{1}{2}\left( {\frac{{{\pi ^2}}}{8} - I} \right)\\ = 2 - \frac{{{\pi ^2}}}{{16}} + \frac{I}{2}\end{array}\]

Xét tích phân\[I = \mathop \smallint \limits_0^{\frac{\pi }{2}} x\cos 2xdx\]

Đặt\(\left\{ {\begin{array}{*{20}{c}}{u = x}\\{dv = cos2xdx}\end{array}} \right. \Leftrightarrow \left\{ {\begin{array}{*{20}{c}}{du = dx}\\{v = \frac{{sin2x}}{2}}\end{array}} \right.\) khi đó ta có:

\[I = x.\frac{{sin2x}}{2}\left| {_0^{\frac{\pi }{2}}} \right. - \frac{1}{2}\int\limits_0^{\frac{\pi }{2}} {sin2xdx} \]

\[I = \frac{\pi }{2}.\frac{{sin\pi }}{2} - 0 + \frac{1}{2}.\frac{{cos2x}}{2}\left| {_0^{\frac{\pi }{2}}} \right.\]

\[I = \frac{1}{4}(cos\pi - cos0)\]

\[I = \frac{1}{4}( - 1 - 1) = - \frac{1}{2}\]

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

\[ \Rightarrow a = 7,\,\,b = 4,\,\,c = 16\]

Vậy\[a + b + c = 7 + 4 + 16 = 27\]Đáp án cần chọn là: D