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Câu hỏi: Cho hàm số $f\left( x \right)$ có đạo hàm ${f}'\left( x \right)=\dfrac{1}{{{\left( \sin x+2\cos x \right)}^{2}}}$ với mọi $x\in \left[ 0; \dfrac{\pi }{2} \right]$ và $f\left( 0 \right)=0$. Tích phân $\int\limits_{0}^{\dfrac{\pi }{2}}{f\left( x \right)} \text{d}x$ bằng
A. $\dfrac{3\pi +2\ln 2}{10}$.
B. $\dfrac{\pi -\ln 2}{5}$.
C. $\dfrac{-\pi +\ln 2}{5}$.
D. $\dfrac{\pi +4\ln 2}{20}$.
A. $\dfrac{3\pi +2\ln 2}{10}$.
B. $\dfrac{\pi -\ln 2}{5}$.
C. $\dfrac{-\pi +\ln 2}{5}$.
D. $\dfrac{\pi +4\ln 2}{20}$.
Ta có: $f\left( x \right)=\int{\dfrac{\text{d}x}{{{\left( \sin x+2\cos x \right)}^{2}}}}=\dfrac{1}{5}\int{\dfrac{\text{d}x}{{{\sin }^{2}}\left( x+\alpha \right)}}$, với $\left\{ \begin{aligned}
& \cos \alpha =\dfrac{1}{\sqrt{5}} \\
& \sin \alpha =\dfrac{2}{\sqrt{5}} \\
\end{aligned} \right.\Rightarrow \cot \alpha =\dfrac{1}{2}$.
Từ đó suy ra: $f\left( x \right)=-\dfrac{1}{5}\cot \left( x+\alpha \right)+C$, mà $f\left( 0 \right)=0\Rightarrow C=\dfrac{1}{10}$.
$\int\limits_{0}^{\dfrac{\pi }{2}}{f\left( x \right)} \text{d}x=\left( -\dfrac{1}{5}\ln \left| \sin \left( x+\alpha \right) \right|+\dfrac{x}{10} \right)\mathop{|}_{0}^{\dfrac{\pi }{2}}$ = $-\dfrac{1}{5}\ln \left( \cot \alpha \right)+\dfrac{\pi }{20}=\dfrac{4\ln 2+\pi }{20}$.
& \cos \alpha =\dfrac{1}{\sqrt{5}} \\
& \sin \alpha =\dfrac{2}{\sqrt{5}} \\
\end{aligned} \right.\Rightarrow \cot \alpha =\dfrac{1}{2}$.
Từ đó suy ra: $f\left( x \right)=-\dfrac{1}{5}\cot \left( x+\alpha \right)+C$, mà $f\left( 0 \right)=0\Rightarrow C=\dfrac{1}{10}$.
$\int\limits_{0}^{\dfrac{\pi }{2}}{f\left( x \right)} \text{d}x=\left( -\dfrac{1}{5}\ln \left| \sin \left( x+\alpha \right) \right|+\dfrac{x}{10} \right)\mathop{|}_{0}^{\dfrac{\pi }{2}}$ = $-\dfrac{1}{5}\ln \left( \cot \alpha \right)+\dfrac{\pi }{20}=\dfrac{4\ln 2+\pi }{20}$.
Đáp án D.
