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Câu hỏi: Cho $\int\limits_{0}^{\dfrac{\pi }{2}}{\left( {{\text{e}}^{\cos x}}+\sin x \right)}\sin x \text{d}x= a + b\text{e} + c\pi $. Khi đó giá trị $a+b+c$ là
A. $\dfrac{6}{5}$.
B. $\dfrac{2}{3}$.
C. $\dfrac{1}{4}$.
D. $\dfrac{3}{5}$.
A. $\dfrac{6}{5}$.
B. $\dfrac{2}{3}$.
C. $\dfrac{1}{4}$.
D. $\dfrac{3}{5}$.
Ta có:
$\int\limits_{0}^{\dfrac{\pi }{2}}{\left( {{\text{e}}^{\cos x}}+\sin x \right)\sin x \text{d}x }=\int\limits_{0}^{\dfrac{\pi }{2}}{{{\text{e}}^{\cos x}}\sin x \text{d}x }+\int\limits_{0}^{\dfrac{\pi }{2}}{{{\sin }^{2}}x \text{d}x }=-\int\limits_{0}^{\dfrac{\pi }{2}}{{{\text{e}}^{\cos x}}\text{d}\left( \cos x \right)}+\dfrac{1}{2}\int\limits_{0}^{\dfrac{\pi }{2}}{\left( 1- \cos 2x \right)\text{d}x}$
$=\left. -{{\text{e}}^{\cos x}} \right|_{0}^{\dfrac{\pi }{2}}\text{+}\left. \dfrac{1}{2}\left( x-\dfrac{\sin 2x}{2} \right) \right|_{0}^{\dfrac{\pi }{2}}\text{=} \text{e}- \text{1} \text{+} \dfrac{\text{1}}{\text{2}}\left( \dfrac{\!\!\pi\!\!}{\text{2}} \right)\text{=} - \text{1} \text{+} \text{e} \text{+} \dfrac{\!\!\pi\!\!}{\text{4}}$
Vậy $a+b+c=\dfrac{1}{4}.$
$\int\limits_{0}^{\dfrac{\pi }{2}}{\left( {{\text{e}}^{\cos x}}+\sin x \right)\sin x \text{d}x }=\int\limits_{0}^{\dfrac{\pi }{2}}{{{\text{e}}^{\cos x}}\sin x \text{d}x }+\int\limits_{0}^{\dfrac{\pi }{2}}{{{\sin }^{2}}x \text{d}x }=-\int\limits_{0}^{\dfrac{\pi }{2}}{{{\text{e}}^{\cos x}}\text{d}\left( \cos x \right)}+\dfrac{1}{2}\int\limits_{0}^{\dfrac{\pi }{2}}{\left( 1- \cos 2x \right)\text{d}x}$
$=\left. -{{\text{e}}^{\cos x}} \right|_{0}^{\dfrac{\pi }{2}}\text{+}\left. \dfrac{1}{2}\left( x-\dfrac{\sin 2x}{2} \right) \right|_{0}^{\dfrac{\pi }{2}}\text{=} \text{e}- \text{1} \text{+} \dfrac{\text{1}}{\text{2}}\left( \dfrac{\!\!\pi\!\!}{\text{2}} \right)\text{=} - \text{1} \text{+} \text{e} \text{+} \dfrac{\!\!\pi\!\!}{\text{4}}$
Vậy $a+b+c=\dfrac{1}{4}.$
Đáp án C.