Biết \(\int\limits_{ - \frac{\pi }{6}}^{\frac{\pi }{6}} {\frac{{x\cos x}}{{\sqrt {1 + {x^2}}  + x}}{\rm{d}}x}  = a + \frac{{{\pi ^2}}}{b}

* Hướng dẫn giải

Ta có \(I = \int\limits_{ - \frac{\pi }{6}}^{\frac{\pi }{6}} {\frac{{x\cos x}}{{\sqrt {1 + {x^2}}  + x}}{\rm{d}}x}  = \int\limits_{ - \frac{\pi }{6}}^{\frac{\pi }{6}} {x\cos x\left( {\sqrt {1 + {x^2}}  - x} \right){\rm{d}}x}  = \int\limits_{ - \frac{\pi }{6}}^{\frac{\pi }{6}} {x\left( {\sqrt {1 + {x^2}}  - x} \right)\cos x{\rm{d}}x} .\)

Lại có \(I = \int\limits_{ - \frac{\pi }{6}}^{\frac{\pi }{6}} {\frac{{x\cos x}}{{\sqrt {1 + {x^2}}  + x}}{\rm{d}}x} \mathop  = \limits^{x =  - t} \int\limits_{\frac{\pi }{6}}^{ - \frac{\pi }{6}} {\frac{{\left( { - t} \right)\cos \left( { - t} \right)}}{{\sqrt {1 + {{\left( { - t} \right)}^2}}  - t}}{\rm{d}}\left( { - t} \right) = \int\limits_{\frac{\pi }{6}}^{ - \frac{\pi }{6}} {\frac{{t\cos t}}{{\sqrt {1 + {t^2}}  - t}}{\rm{d}}t} } \)

\( =  - \int\limits_{ - \frac{\pi }{6}}^{\frac{\pi }{6}} {t\left( {\sqrt {1 + {t^2}}  + t} \right)\cos t{\rm{d}}t}  =  - \int\limits_{ - \frac{\pi }{6}}^{\frac{\pi }{6}} {x\left( {\sqrt {1 + {x^2}}  + x} \right)\cos x{\rm{d}}x} .\)

Suy ra \(2I = \int\limits_{ - \frac{\pi }{6}}^{\frac{\pi }{6}} {x\left( {\sqrt {1 + {x^2}}  - x} \right)\cos x{\rm{d}}x}  - \int\limits_{ - \frac{\pi }{6}}^{\frac{\pi }{6}} {x\left( {\sqrt {1 + {x^2}}  + x} \right)\cos x{\rm{d}}x}  =  - 2\int\limits_{ - \frac{\pi }{6}}^{\frac{\pi }{6}} {{x^2}\cos x{\rm{d}}x} \)

\( \to I =  - \int\limits_{ - \frac{\pi }{6}}^{\frac{\pi }{6}} {{x^2}\cos x{\rm{d}}x} .\) Tích phân từng phần hai lần ta được \(I = 2 + \frac{{{\pi ^2}}}{{ - 36}} + \frac{{\sqrt 3 \pi }}{{ - 3}}\)

\( \to \left\{ \begin{array}{l}
a = 2\\
b =  - 36\\
c =  - 3
\end{array} \right. \to P = a - b + c = 35.\)