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Câu hỏi: Biết $\int\limits_{\dfrac{\pi }{4}}^{\dfrac{\pi }{2}}{\dfrac{x}{{{\sin }^{2}}x}dx=m\pi +n\ln 2\left( m,n\in \mathbb{R} \right)}$, hãy tính giá trị biểu thức $P=2m+n$.
A. P = 0,25.
B. P = 0,75.
C. P = 1.
D. P = 0.
A. P = 0,25.
B. P = 0,75.
C. P = 1.
D. P = 0.
Chọn $\left\{ \begin{aligned}
& u=x \\
& dv=\dfrac{1}{{{\sin }^{2}}x} \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& du=dx \\
& v=-\cot x \\
\end{aligned} \right.$
Khi đó $\int\limits_{\dfrac{\pi }{4}}^{\dfrac{\pi }{2}}{\dfrac{x}{{{\sin }^{2}}x}dx=-x\cot x}\mathop{|}_{\dfrac{\pi }{4}}^{\dfrac{\pi }{2}}+\int\limits_{\dfrac{\pi }{4}}^{\dfrac{\pi }{2}}{\cot xdx}=\dfrac{\pi }{4}+\ln \left| \sin x \right|\mathop{|}_{\dfrac{\pi }{4}}^{\dfrac{\pi }{2}}=\dfrac{\pi }{4}+\dfrac{1}{2}\ln 2$
Đồng nhất hai vế $\dfrac{\pi }{4}+\dfrac{1}{2}\ln 2=m\pi +n\ln 2\Rightarrow m=\dfrac{1}{4},n=\dfrac{1}{2}\Rightarrow P=2m+n=1.$
& u=x \\
& dv=\dfrac{1}{{{\sin }^{2}}x} \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& du=dx \\
& v=-\cot x \\
\end{aligned} \right.$
Khi đó $\int\limits_{\dfrac{\pi }{4}}^{\dfrac{\pi }{2}}{\dfrac{x}{{{\sin }^{2}}x}dx=-x\cot x}\mathop{|}_{\dfrac{\pi }{4}}^{\dfrac{\pi }{2}}+\int\limits_{\dfrac{\pi }{4}}^{\dfrac{\pi }{2}}{\cot xdx}=\dfrac{\pi }{4}+\ln \left| \sin x \right|\mathop{|}_{\dfrac{\pi }{4}}^{\dfrac{\pi }{2}}=\dfrac{\pi }{4}+\dfrac{1}{2}\ln 2$
Đồng nhất hai vế $\dfrac{\pi }{4}+\dfrac{1}{2}\ln 2=m\pi +n\ln 2\Rightarrow m=\dfrac{1}{4},n=\dfrac{1}{2}\Rightarrow P=2m+n=1.$
Đáp án C.