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Câu hỏi: Họ nguyên hàm của hàm số $f\left( x \right)=\dfrac{x}{{{\sin }^{2}}x}$ trên khoảng $\left( 0;\pi \right)$ là
A. $-x\cot x+\ln \left( \sin x \right)+C$.
B. $x\cot x-\ln \left| \sin x \right|+C$.
C. $x\cot x+\ln \left| \sin x \right|+C$.
D. $-x\cot x-\ln \left( \sin x \right)+C$.
A. $-x\cot x+\ln \left( \sin x \right)+C$.
B. $x\cot x-\ln \left| \sin x \right|+C$.
C. $x\cot x+\ln \left| \sin x \right|+C$.
D. $-x\cot x-\ln \left( \sin x \right)+C$.
Đặt $\left\{ \begin{aligned}
& u=x \\
& dv=\dfrac{dx}{{{\sin }^{2}}x} \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& du=dx \\
& v=-\cot x \\
\end{aligned} \right.$
Suy ra $\int{f\left( x \right)dx}=-x.\cot x+\int{\cot xdx}$
$\Rightarrow \int{f\left( x \right)dx}=-x.\cot x+\int{\dfrac{\cos x}{\sin x}dx}=-x.\cot x+\int{\dfrac{d\left( \sin x \right)}{\sin x}}$
$=-x.\cot x+\ln \left| \sin x \right|+C=-x.\cot x+\ln \left( \sin x \right)+C$
Vì $\left| \sin x \right|=\sin x$ khi $0<x<\pi $.
& u=x \\
& dv=\dfrac{dx}{{{\sin }^{2}}x} \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& du=dx \\
& v=-\cot x \\
\end{aligned} \right.$
Suy ra $\int{f\left( x \right)dx}=-x.\cot x+\int{\cot xdx}$
$\Rightarrow \int{f\left( x \right)dx}=-x.\cot x+\int{\dfrac{\cos x}{\sin x}dx}=-x.\cot x+\int{\dfrac{d\left( \sin x \right)}{\sin x}}$
$=-x.\cot x+\ln \left| \sin x \right|+C=-x.\cot x+\ln \left( \sin x \right)+C$
Vì $\left| \sin x \right|=\sin x$ khi $0<x<\pi $.
Đáp án A.