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Câu hỏi: Biết $I=\int\limits_{0}^{\dfrac{\!\!\pi\!\!}{2}}{\dfrac{x+x\cos x-{{\sin }^{3}}x}{1+\cos x}\text{d}x}=\dfrac{{{\!\!\pi\!\!}^{2}}}{a}-\dfrac{b}{c}$. Trong đó $a$, $b$, $z+{{\left| z \right|}^{2}}.i-1-\dfrac{3}{4}i=0$ là các số nguyên dương, phân số $\dfrac{b}{c}$ tối giản. Tính $T={{a}^{2}}+{{b}^{2}}+{{c}^{2}}$.
A. $T=50$.
B. $T=59$.
C. $T=16$.
D. $T=69$.
A. $T=50$.
B. $T=59$.
C. $T=16$.
D. $T=69$.
$I=\int\limits_{0}^{\dfrac{\!\!\pi\!\!}{2}}{\dfrac{x+x\cos x-{{\sin }^{3}}x}{1+\cos x}\text{d}x}=\int\limits_{0}^{\dfrac{\pi }{2}}{\left( x-\dfrac{1}{2}\sin 2x \right)}dx=\left( \dfrac{{{x}^{2}}}{2}+\dfrac{1}{4}\cos 2x \right)\left| \begin{aligned}
& \dfrac{\pi }{2} \\
& 0 \\
\end{aligned} \right.=\dfrac{{{\pi }^{2}}}{8}-\dfrac{1}{2}$.
$\Rightarrow \dfrac{{{\pi }^{2}}}{a}-\dfrac{b}{c}=\dfrac{{{\pi }^{2}}}{8}-\dfrac{1}{2}$.
$\Rightarrow \left\{ \begin{aligned}
& a=8 \\
& b=1 \\
& c=2 \\
\end{aligned} \right.\Rightarrow {{a}^{2}}+{{b}^{2}}+{{c}^{2}}=69$.
& \dfrac{\pi }{2} \\
& 0 \\
\end{aligned} \right.=\dfrac{{{\pi }^{2}}}{8}-\dfrac{1}{2}$.
$\Rightarrow \dfrac{{{\pi }^{2}}}{a}-\dfrac{b}{c}=\dfrac{{{\pi }^{2}}}{8}-\dfrac{1}{2}$.
$\Rightarrow \left\{ \begin{aligned}
& a=8 \\
& b=1 \\
& c=2 \\
\end{aligned} \right.\Rightarrow {{a}^{2}}+{{b}^{2}}+{{c}^{2}}=69$.
Đáp án D.