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Câu hỏi: Cho hàm số $f(x)$ liên tục trên $R$, biết $\dfrac{1}{2} f(1-\sin x)+\sin x f\left(2-\cos ^{2} x\right)=\dfrac{\sin ^{2} x}{\sin x+2}$. Khi đó $\int_{0}^{2} f(x) d x$ là
A. $\dfrac{3}{2}+4 \ln \dfrac{3}{2}$
B. $-3+8 \ln \dfrac{3}{2}$.
C. $3-8 \ln \dfrac{3}{2}$
D. $\dfrac{3}{2}-4 \ln \dfrac{3}{2}$.
A. $\dfrac{3}{2}+4 \ln \dfrac{3}{2}$
B. $-3+8 \ln \dfrac{3}{2}$.
C. $3-8 \ln \dfrac{3}{2}$
D. $\dfrac{3}{2}-4 \ln \dfrac{3}{2}$.
$\begin{aligned}
& \dfrac{1}{2}f(1-\sin x)+\sin x.f\left( 2-{{\cos }^{2}}x \right)=\dfrac{{{\sin }^{2}}x}{\sin x+2} \\
& \Rightarrow \cos x.f(1-\sin x)+2\sin x.\cos xf\left( 2-{{\cos }^{2}}x \right)=\dfrac{2{{\sin }^{2}}x.\cos x}{\sin x+2} \\
\end{aligned}$
$\Rightarrow \int\limits_{0}^{\dfrac{\pi }{2}}{\left[ \cos x.f(1-\sin x)+2\sin x.\cos x.f\left( 2-{{\cos }^{2}}x \right) \right]}dx=\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{2{{\sin }^{2}}x.\cos x}{\sin x+2}dx}$
$\Leftrightarrow \int\limits_{0}^{\dfrac{\pi }{2}}{\cos x.f(1-\sin x)dx+\int\limits_{0}^{\dfrac{\pi }{2}}{f\left( 2-{{\cos }^{2}}x \right).\sin 2x.dx}}=\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{2{{\sin }^{2}}x.\cos x}{\sin x+2}dx}$
+ Đặt ${{I}_{1}}=\int\limits_{0}^{\dfrac{\pi }{2}}{\cos x.f(1-\sin x)dx}$. Đặt $t=1-\sin x\Rightarrow dt=-\cos xdx$.
Đổi cận $\left\{ \begin{aligned}
& x=0 \\
& x=\dfrac{\pi }{2} \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& t=1 \\
& t=0 \\
\end{aligned} \right.$$\Rightarrow {{I}_{1}}=\int\limits_{0}^{1}{f(t)dt=}\int\limits_{0}^{1}{f(x)dx}$
+ Đặt ${{I}_{2}}=\int\limits_{0}^{\dfrac{\pi }{2}}{f(2-{{\cos }^{2}}x).\sin 2xdx}$. Đặt $t=2-{{\cos }^{2}}x\Rightarrow dt=\sin 2x.dx$.
Đổi cận $\left\{ \begin{aligned}
& x=0 \\
& x=\dfrac{\pi }{2} \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& t=1 \\
& t=2 \\
\end{aligned} \right.$$\Rightarrow {{I}_{2}}=\int\limits_{1}^{2}{f(t)dt=}\int\limits_{1}^{2}{f(x)dx}$
+ Đặt ${{I}_{3}}=\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{{{\sin }^{2}}x}{\sin x+2}.\cos x.dx}$. Đặt $t=2+\sin x\Rightarrow dt=\cos x.dx$.
Đổi cận $\left\{ \begin{aligned}
& x=0 \\
& x=\dfrac{\pi }{2} \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& t=2 \\
& t=3 \\
\end{aligned} \right.$$\Rightarrow {{I}_{3}}=\int\limits_{2}^{3}{\dfrac{{{\left( t-2 \right)}^{2}}}{t}}dt=\int\limits_{2}^{3}{\left( t-4+\dfrac{4}{t} \right)}dt=\left( \dfrac{{{t}^{2}}}{2}-4t+4\ln t \right)|_{2}^{3}=\dfrac{-3}{2}+4\ln \dfrac{3}{2}$
Do đó $\int_{0}^{2}{f}(x)dx={{I}_{1}}+{{I}_{2}}=-3+8\ln \dfrac{3}{2}$.
& \dfrac{1}{2}f(1-\sin x)+\sin x.f\left( 2-{{\cos }^{2}}x \right)=\dfrac{{{\sin }^{2}}x}{\sin x+2} \\
& \Rightarrow \cos x.f(1-\sin x)+2\sin x.\cos xf\left( 2-{{\cos }^{2}}x \right)=\dfrac{2{{\sin }^{2}}x.\cos x}{\sin x+2} \\
\end{aligned}$
$\Rightarrow \int\limits_{0}^{\dfrac{\pi }{2}}{\left[ \cos x.f(1-\sin x)+2\sin x.\cos x.f\left( 2-{{\cos }^{2}}x \right) \right]}dx=\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{2{{\sin }^{2}}x.\cos x}{\sin x+2}dx}$
$\Leftrightarrow \int\limits_{0}^{\dfrac{\pi }{2}}{\cos x.f(1-\sin x)dx+\int\limits_{0}^{\dfrac{\pi }{2}}{f\left( 2-{{\cos }^{2}}x \right).\sin 2x.dx}}=\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{2{{\sin }^{2}}x.\cos x}{\sin x+2}dx}$
+ Đặt ${{I}_{1}}=\int\limits_{0}^{\dfrac{\pi }{2}}{\cos x.f(1-\sin x)dx}$. Đặt $t=1-\sin x\Rightarrow dt=-\cos xdx$.
Đổi cận $\left\{ \begin{aligned}
& x=0 \\
& x=\dfrac{\pi }{2} \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& t=1 \\
& t=0 \\
\end{aligned} \right.$$\Rightarrow {{I}_{1}}=\int\limits_{0}^{1}{f(t)dt=}\int\limits_{0}^{1}{f(x)dx}$
+ Đặt ${{I}_{2}}=\int\limits_{0}^{\dfrac{\pi }{2}}{f(2-{{\cos }^{2}}x).\sin 2xdx}$. Đặt $t=2-{{\cos }^{2}}x\Rightarrow dt=\sin 2x.dx$.
Đổi cận $\left\{ \begin{aligned}
& x=0 \\
& x=\dfrac{\pi }{2} \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& t=1 \\
& t=2 \\
\end{aligned} \right.$$\Rightarrow {{I}_{2}}=\int\limits_{1}^{2}{f(t)dt=}\int\limits_{1}^{2}{f(x)dx}$
+ Đặt ${{I}_{3}}=\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{{{\sin }^{2}}x}{\sin x+2}.\cos x.dx}$. Đặt $t=2+\sin x\Rightarrow dt=\cos x.dx$.
Đổi cận $\left\{ \begin{aligned}
& x=0 \\
& x=\dfrac{\pi }{2} \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& t=2 \\
& t=3 \\
\end{aligned} \right.$$\Rightarrow {{I}_{3}}=\int\limits_{2}^{3}{\dfrac{{{\left( t-2 \right)}^{2}}}{t}}dt=\int\limits_{2}^{3}{\left( t-4+\dfrac{4}{t} \right)}dt=\left( \dfrac{{{t}^{2}}}{2}-4t+4\ln t \right)|_{2}^{3}=\dfrac{-3}{2}+4\ln \dfrac{3}{2}$
Do đó $\int_{0}^{2}{f}(x)dx={{I}_{1}}+{{I}_{2}}=-3+8\ln \dfrac{3}{2}$.
Đáp án B.