Google
Câu hỏi: Cho hàm số $f\left( x \right)$ có $f\left( \dfrac{\pi }{2} \right)=2$ và $f'\left( x \right)=x\sin x$.
Giả sử rằng $\int\limits_{0}^{\dfrac{\pi }{2}}{\cos x.f\left( x \right)}\text{dx=}\dfrac{a}{b}-\dfrac{{{\pi }^{2}}}{c}$ ( với $a,b,c$ là các số nguyên dương, $\dfrac{a}{b}$ tối giản). Khi đó $a+b+c$ bằng
A. $23$.
B. $5$.
C. $20$.
D. $27$.
Giả sử rằng $\int\limits_{0}^{\dfrac{\pi }{2}}{\cos x.f\left( x \right)}\text{dx=}\dfrac{a}{b}-\dfrac{{{\pi }^{2}}}{c}$ ( với $a,b,c$ là các số nguyên dương, $\dfrac{a}{b}$ tối giản). Khi đó $a+b+c$ bằng
A. $23$.
B. $5$.
C. $20$.
D. $27$.
Do $f'\left( x \right)=x\sin x$ nên $f\left( x \right)=\int{f'\left( x \right)}\text{dx=}\int{x\sin x\text{dx}}$ $=-\int{x.\text{d}\left( \cos x \right)}=-x.\cos x+\int{\cos x\text{dx}}$
$=-x\cos x+\sin x+C$.
Theo giả thiết $f\left( \dfrac{\pi }{2} \right)=2$ $\Leftrightarrow -\dfrac{\pi }{2}\cos \dfrac{\pi }{2}+\sin \dfrac{\pi }{2}+C=2\Leftrightarrow C=1$.
Vậy $f\left( x \right)=-x\cos x+\sin x+1$.
$\int\limits_{0}^{\dfrac{\pi }{2}}{\cos x.f\left( x \right)}\text{dx=}\int\limits_{0}^{\dfrac{\pi }{2}}{\cos x.\left( -x\cos x+\sin x+1 \right)}\text{dx=}\int\limits_{0}^{\dfrac{\pi }{2}}{\left( -x.{{\cos }^{2}}x+\sin x.\cos x+\cos x \right)}\text{dx}$
$=-\int\limits_{0}^{\dfrac{\pi }{2}}{x.{{\cos }^{2}}x\text{dx} \text{+ }}\dfrac{1}{2}\int\limits_{0}^{\dfrac{\pi }{2}}{\sin 2x\text{dx} \text{+ }}\int\limits_{0}^{\dfrac{\pi }{2}}{\cos x\text{dx} }$ $=-\dfrac{1}{2}\int\limits_{0}^{\dfrac{\pi }{2}}{x\left( 1+\cos 2x \right)\text{dx+}\dfrac{1}{2}.\dfrac{1}{2}\int\limits_{0}^{\dfrac{\pi }{2}}{\sin 2x\text{d}\left( 2x \right)}}+\int\limits_{0}^{\dfrac{\pi }{2}}{\cos x\text{dx}}$
$=-\dfrac{1}{2}\int\limits_{0}^{\dfrac{\pi }{2}}{x\text{dx}}-\dfrac{1}{2}\int\limits_{0}^{\dfrac{\pi }{2}}{x.\cos 2x\text{dx}+}\dfrac{1}{4}.\left. \left( -\cos 2x \right) \right|_{0}^{\dfrac{\pi }{2}}+\left. \sin x \right|_{0}^{\dfrac{\pi }{2}}$
$=\dfrac{-1}{2}.\dfrac{{{x}^{2}}}{2}\left| \begin{aligned}
& \dfrac{\pi }{2} \\
& 0 \\
\end{aligned} \right.-\dfrac{1}{4}\cos 2x\left| \begin{aligned}
& \dfrac{\pi }{2} \\
& 0 \\
\end{aligned} \right.+\sin x\left| \begin{aligned}
& \dfrac{\pi }{2} \\
& 0 \\
\end{aligned} \right.-\dfrac{1}{2}\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{1}{2}x}.d(\sin 2x)$
$=-\dfrac{{{\pi }^{2}}}{16}+\dfrac{1}{2}+1-\dfrac{1}{4}x.\left. \sin 2x \right|_{0}^{\dfrac{\pi }{2}}+\dfrac{1}{4}\int\limits_{0}^{\dfrac{\pi }{2}}{\sin 2x\text{dx}}$
$=-\dfrac{{{\pi }^{2}}}{16}+\dfrac{3}{2}-\dfrac{1}{8}\left. \cos 2x \right|_{0}^{\dfrac{\pi }{2}}=\dfrac{7}{4}-\dfrac{{{\pi }^{2}}}{16}$.
Vậy $a=7;\ b=4;\ c=16$. Khi đó: $a+b+c=7+4+16=27$.
$=-x\cos x+\sin x+C$.
Theo giả thiết $f\left( \dfrac{\pi }{2} \right)=2$ $\Leftrightarrow -\dfrac{\pi }{2}\cos \dfrac{\pi }{2}+\sin \dfrac{\pi }{2}+C=2\Leftrightarrow C=1$.
Vậy $f\left( x \right)=-x\cos x+\sin x+1$.
$\int\limits_{0}^{\dfrac{\pi }{2}}{\cos x.f\left( x \right)}\text{dx=}\int\limits_{0}^{\dfrac{\pi }{2}}{\cos x.\left( -x\cos x+\sin x+1 \right)}\text{dx=}\int\limits_{0}^{\dfrac{\pi }{2}}{\left( -x.{{\cos }^{2}}x+\sin x.\cos x+\cos x \right)}\text{dx}$
$=-\int\limits_{0}^{\dfrac{\pi }{2}}{x.{{\cos }^{2}}x\text{dx} \text{+ }}\dfrac{1}{2}\int\limits_{0}^{\dfrac{\pi }{2}}{\sin 2x\text{dx} \text{+ }}\int\limits_{0}^{\dfrac{\pi }{2}}{\cos x\text{dx} }$ $=-\dfrac{1}{2}\int\limits_{0}^{\dfrac{\pi }{2}}{x\left( 1+\cos 2x \right)\text{dx+}\dfrac{1}{2}.\dfrac{1}{2}\int\limits_{0}^{\dfrac{\pi }{2}}{\sin 2x\text{d}\left( 2x \right)}}+\int\limits_{0}^{\dfrac{\pi }{2}}{\cos x\text{dx}}$
$=-\dfrac{1}{2}\int\limits_{0}^{\dfrac{\pi }{2}}{x\text{dx}}-\dfrac{1}{2}\int\limits_{0}^{\dfrac{\pi }{2}}{x.\cos 2x\text{dx}+}\dfrac{1}{4}.\left. \left( -\cos 2x \right) \right|_{0}^{\dfrac{\pi }{2}}+\left. \sin x \right|_{0}^{\dfrac{\pi }{2}}$
$=\dfrac{-1}{2}.\dfrac{{{x}^{2}}}{2}\left| \begin{aligned}
& \dfrac{\pi }{2} \\
& 0 \\
\end{aligned} \right.-\dfrac{1}{4}\cos 2x\left| \begin{aligned}
& \dfrac{\pi }{2} \\
& 0 \\
\end{aligned} \right.+\sin x\left| \begin{aligned}
& \dfrac{\pi }{2} \\
& 0 \\
\end{aligned} \right.-\dfrac{1}{2}\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{1}{2}x}.d(\sin 2x)$
$=-\dfrac{{{\pi }^{2}}}{16}+\dfrac{1}{2}+1-\dfrac{1}{4}x.\left. \sin 2x \right|_{0}^{\dfrac{\pi }{2}}+\dfrac{1}{4}\int\limits_{0}^{\dfrac{\pi }{2}}{\sin 2x\text{dx}}$
$=-\dfrac{{{\pi }^{2}}}{16}+\dfrac{3}{2}-\dfrac{1}{8}\left. \cos 2x \right|_{0}^{\dfrac{\pi }{2}}=\dfrac{7}{4}-\dfrac{{{\pi }^{2}}}{16}$.
Vậy $a=7;\ b=4;\ c=16$. Khi đó: $a+b+c=7+4+16=27$.
Đáp án D.