Biết \(\int\limits_0^{\frac{\pi }{2}} {\frac{{{x^2} + \left( {2x + \cos x} \right)\cos x + 1 - \sin x}}{{x + \cos x}}{\rm{d}}x}  = a{\pi ^2} +

* Hướng dẫn giải

Ta có \(I = \int\limits_0^{\frac{\pi }{2}} {\frac{{\left( {{x^2} + 2x\cos x + {{\cos }^2}x} \right) + \left( {1 - \sin x} \right)}}{{x + \cos x}}{\rm{d}}x} \)

\(\begin{array}{l}
 = \int\limits_0^{\frac{\pi }{2}} {\frac{{{{\left( {x + \cos x} \right)}^2}}}{{x + \cos x}}{\rm{d}}x}  + \int\limits_0^{\frac{\pi }{2}} {\frac{{1 - \sin x}}{{x + \cos x}}{\rm{d}}x}  = \int\limits_0^{\frac{\pi }{2}} {\left( {x + \cos x} \right){\rm{d}}x}  + \int\limits_0^{\frac{\pi }{2}} {\frac{{{\rm{d}}\left( {x + \cos x} \right)}}{{x + \cos x}}} \\
 = \left( {\frac{1}{2}{x^2} + \sin x + \ln \left| {x + \cos x} \right|} \right)\left| \begin{array}{l}
^{\frac{\pi }{2}}\\
_0
\end{array} \right. = \frac{1}{8}{\pi ^2} + 1 + \ln \frac{\pi }{2} = \frac{1}{8}{\pi ^2} + 1 - \ln \frac{2}{\pi }\\
 \to \left\{ \begin{array}{l}
a = \frac{1}{8}\\
b = 1\\
c = 2
\end{array} \right. \to P = a{c^3} + b = 2.
\end{array}\)