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Câu hỏi: Cho hàm số f(x) liên tục trên đoạn $\left[ \dfrac{\pi }{4};\dfrac{\pi }{3} \right]$ thỏa mãn ${f}'\left( x \right).\sin 2x=1+2.f\left( x \right)$ với $\forall x\in \left[ \dfrac{\pi }{4};\dfrac{\pi }{3} \right]$ và $f\left( \dfrac{\pi }{4} \right)=1.$ Tích phân $I=\int\limits_{0}^{\dfrac{\pi }{3}}{f\left( x \right)dx}$ bằng
A. $-\dfrac{\pi }{24}+\dfrac{3}{4}\ln 2.$
B. $-\dfrac{\pi }{24}+\dfrac{1}{4}\ln 2.$
C. $-\dfrac{\pi }{12}+\dfrac{3}{2}\ln 2.$
D. $-\dfrac{\pi }{12}+\dfrac{1}{8}\ln 2.$
A. $-\dfrac{\pi }{24}+\dfrac{3}{4}\ln 2.$
B. $-\dfrac{\pi }{24}+\dfrac{1}{4}\ln 2.$
C. $-\dfrac{\pi }{12}+\dfrac{3}{2}\ln 2.$
D. $-\dfrac{\pi }{12}+\dfrac{1}{8}\ln 2.$
Ta có
${{\left[ \dfrac{f\left( x \right)}{\tan x} \right]}^{\prime }}=\dfrac{{f}'\left( x \right).\tan x-f\left( x \right).\dfrac{1}{{{\cos }^{2}}x}}{{{\tan }^{2}}x}=\dfrac{{f}'\left( x \right).\sin x\cos x-f\left( x \right)}{{{\sin }^{2}}x}$
$=\dfrac{{f}'\left( x \right).\sin 2x-2.f\left( x \right)}{2{{\sin }^{2}}x}=\dfrac{1}{2{{\sin }^{2}}x}\Rightarrow \dfrac{f\left( x \right)}{\tan x}=\mathop{\int }^{}\dfrac{1}{2{{\sin }^{2}}x}dx=-\dfrac{1}{2}\cot x+C.$
Bài ra
$f\left( \dfrac{\pi }{4} \right)=1\Rightarrow C=\dfrac{3}{2}\Rightarrow f\left( x \right)=\tan x\left( -\dfrac{1}{2}\cot x+\dfrac{3}{2} \right)=-\dfrac{1}{2}+\dfrac{3}{2}.\dfrac{\sin x}{\cos x}$
$\Rightarrow \int\limits_{\dfrac{\pi }{4}}^{\dfrac{\pi }{3}}{f\left( x \right)dx}=\left( -\dfrac{1}{2}x-\dfrac{3}{2}\ln \left| \cos x \right| \right)\left| _{\begin{smallmatrix}
\\
\dfrac{\pi }{4}
\end{smallmatrix}}^{\begin{smallmatrix}
\dfrac{\pi }{3} \\
\end{smallmatrix}} \right.=-\dfrac{\pi }{6}-\dfrac{3}{2}\ln \dfrac{1}{2}+\dfrac{\pi }{8}+\dfrac{3}{2}\ln \dfrac{1}{\sqrt{2}}=-\dfrac{\pi }{24}+\dfrac{3}{4}\ln 2.$
${{\left[ \dfrac{f\left( x \right)}{\tan x} \right]}^{\prime }}=\dfrac{{f}'\left( x \right).\tan x-f\left( x \right).\dfrac{1}{{{\cos }^{2}}x}}{{{\tan }^{2}}x}=\dfrac{{f}'\left( x \right).\sin x\cos x-f\left( x \right)}{{{\sin }^{2}}x}$
$=\dfrac{{f}'\left( x \right).\sin 2x-2.f\left( x \right)}{2{{\sin }^{2}}x}=\dfrac{1}{2{{\sin }^{2}}x}\Rightarrow \dfrac{f\left( x \right)}{\tan x}=\mathop{\int }^{}\dfrac{1}{2{{\sin }^{2}}x}dx=-\dfrac{1}{2}\cot x+C.$
Bài ra
$f\left( \dfrac{\pi }{4} \right)=1\Rightarrow C=\dfrac{3}{2}\Rightarrow f\left( x \right)=\tan x\left( -\dfrac{1}{2}\cot x+\dfrac{3}{2} \right)=-\dfrac{1}{2}+\dfrac{3}{2}.\dfrac{\sin x}{\cos x}$
$\Rightarrow \int\limits_{\dfrac{\pi }{4}}^{\dfrac{\pi }{3}}{f\left( x \right)dx}=\left( -\dfrac{1}{2}x-\dfrac{3}{2}\ln \left| \cos x \right| \right)\left| _{\begin{smallmatrix}
\\
\dfrac{\pi }{4}
\end{smallmatrix}}^{\begin{smallmatrix}
\dfrac{\pi }{3} \\
\end{smallmatrix}} \right.=-\dfrac{\pi }{6}-\dfrac{3}{2}\ln \dfrac{1}{2}+\dfrac{\pi }{8}+\dfrac{3}{2}\ln \dfrac{1}{\sqrt{2}}=-\dfrac{\pi }{24}+\dfrac{3}{4}\ln 2.$
Đáp án A.