Cho biết $\int\limits_{0}^{\pi }{{{\left( x+\cos x...

Ta có $I=\int\limits_{0}^{\pi }{{{\left( x+\cos x \right)}^{2}}\text{d}x}=\int\limits_{0}^{\pi }{\left( {{x}^{2}}+2x\cos x+{{\cos }^{2}}x \right)\text{d}x}=\int\limits_{0}^{\pi }{{{x}^{2}}\text{d}x}+2\int\limits_{0}^{\pi }{x\cos x\text{d}x}+\int\limits_{0}^{\pi }{{{\cos }^{2}}x\text{d}x}$.
Với $A=\int\limits_{0}^{\pi }{{{x}^{2}}\text{d}x}=\dfrac{{{\pi }^{3}}}{3}$.
Với $B=\int\limits_{0}^{\pi }{x\cos x\text{d}x}$ sử dụng từng phần đặt $\left\{ \begin{aligned}
& u=x \\
& \text{d}v=\cos x\text{d}x \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& \text{d}u=\text{d}x \\
& v=\sin x \\
\end{aligned} \right.$.
Suy ra $B=\left. \left( x\sin x \right) \right|_{0}^{\pi }-\int\limits_{0}^{\pi }{\sin x\text{d}x}=\left. \cos x \right|_{0}^{\pi }=-2$
Với $C=\int\limits_{0}^{\pi }{{{\cos }^{2}}x\text{d}x}=\int\limits_{0}^{\pi }{\dfrac{1+\cos 2x}{2}\text{d}x=\left. \left[ \dfrac{x}{2}+\dfrac{\sin 2x}{4} \right] \right|}_{0}^{\pi }=\dfrac{\pi }{2}$.
Suy ra $I=A+2B+C=\dfrac{{{\pi }^{3}}}{3}-4+\dfrac{\pi }{2}\equiv \dfrac{{{\pi }^{3}}}{a}+\dfrac{\pi }{b}-c\Rightarrow a=3, b=2, c=4\Rightarrow T=a+b+c=9$.