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Câu hỏi: Cho hàm số $f(x)$ biết $f(0)=1$ và ${f}'\left( x \right)=\dfrac{1}{1+\sin x},\ \forall x\in \left[ 0;\dfrac{\pi }{2} \right]$. Tích phân $\int\limits_{0}^{\dfrac{\pi }{2}}{f(x)}dx$ bằng
A. $\pi +\dfrac{1}{2}\ln 2$.
B. $\pi -\dfrac{1}{2}\ln 2$.
C. $\pi -\ln 2$.
D. $\pi +\ln 2$.
A. $\pi +\dfrac{1}{2}\ln 2$.
B. $\pi -\dfrac{1}{2}\ln 2$.
C. $\pi -\ln 2$.
D. $\pi +\ln 2$.
$x\in \left[ 0;\dfrac{\pi }{2} \right],f'\left( x \right)=\dfrac{1}{1+\sin x}=\dfrac{1}{{{\left( \sin \dfrac{x}{2}+\cos \dfrac{x}{2} \right)}^{2}}}=\dfrac{1}{{{\left[ \sqrt{2}\cos \left( \dfrac{x}{2}-\dfrac{\pi }{4} \right) \right]}^{2}}}=\dfrac{1}{2{{\cos }^{2}}\left( \dfrac{x}{2}-\dfrac{\pi }{4} \right)}$
Suy ra $f(x)=\int{{f}'\left( x \right)}dx=\int{\dfrac{1}{2{{\cos }^{2}}\left( \dfrac{x}{2}-\dfrac{\pi }{4} \right)}dx=\tan \left( \dfrac{x}{2}-\dfrac{\pi }{4} \right)}+C$.
Lại có $f\left( 0 \right)=1\Rightarrow C=2$
$\Rightarrow $ $\int\limits_{0}^{\dfrac{\pi }{2}}{f(x)}dx=\int\limits_{0}^{\dfrac{\pi }{2}}{\left[ \tan \left( \dfrac{x}{2}-\dfrac{\pi }{4} \right)+2 \right]}dx=-2\left. \left( \ln \left| \cos \left( \dfrac{x}{2}-\dfrac{\pi }{4} \right) \right| \right) \right|_{0}^{\dfrac{\pi }{2}}+\left. \left( 2x \right) \right|_{0}^{\dfrac{\pi }{2}}=\pi -\ln 2.$
Suy ra $f(x)=\int{{f}'\left( x \right)}dx=\int{\dfrac{1}{2{{\cos }^{2}}\left( \dfrac{x}{2}-\dfrac{\pi }{4} \right)}dx=\tan \left( \dfrac{x}{2}-\dfrac{\pi }{4} \right)}+C$.
Lại có $f\left( 0 \right)=1\Rightarrow C=2$
$\Rightarrow $ $\int\limits_{0}^{\dfrac{\pi }{2}}{f(x)}dx=\int\limits_{0}^{\dfrac{\pi }{2}}{\left[ \tan \left( \dfrac{x}{2}-\dfrac{\pi }{4} \right)+2 \right]}dx=-2\left. \left( \ln \left| \cos \left( \dfrac{x}{2}-\dfrac{\pi }{4} \right) \right| \right) \right|_{0}^{\dfrac{\pi }{2}}+\left. \left( 2x \right) \right|_{0}^{\dfrac{\pi }{2}}=\pi -\ln 2.$
Đáp án C.