\(F(x)\) là một nguyên hàm của hàm số \(f(x)=\cot x\) và \(F\left(\dfrac{\pi}{2}\right)=0\). Giá trị của \(F\left(\dfrac{\pi}{6}\right)\) bằng
 | \(-\ln\left(\dfrac{\sqrt{3}}{2}\right)\) |
 | \(\ln\left(\dfrac{\sqrt{3}}{2}\right)\) |
 | \(\ln2\) |
 | \(-\ln2\) |
Chọn phương án D.
$\begin{eqnarray*}
&F(x)\bigg|_{\tfrac{\pi}{2}}^{\tfrac{\pi}{6}}&=\displaystyle\int\limits_{\tfrac{\pi}{2}}^{\tfrac{\pi}{6}}f(x)\mathrm{\,d}x\\
\Leftrightarrow&F\left(\dfrac{\pi}{6}\right)-F\left(\dfrac{\pi}{2}\right)&=\displaystyle\int\limits_{\tfrac{\pi}{2}}^{\tfrac{\pi}{6}}\dfrac{\cos x}{\sin x}\mathrm{\,d}x\\
\Leftrightarrow&F\left(\dfrac{\pi}{6}\right)-0&=\displaystyle\int\limits_{\tfrac{\pi}{2}}^{\tfrac{\pi}{6}}\dfrac{1}{\sin x}\mathrm{\,d}\left(\sin x\right)\\
\Leftrightarrow&F\left(\dfrac{\pi}{6}\right)&=\ln\left|\sin x\right|\bigg|_{\tfrac{\pi}{2}}^{\tfrac{\pi}{6}}\\
\Leftrightarrow&F\left(\dfrac{\pi}{6}\right)&=\ln\dfrac{1}{2}=-\ln2.
\end{eqnarray*}$
Chọn phương án D.
Đặt \(u=\sin x\Rightarrow\mathrm{d}u=\cos x\cdot\mathrm{\,d}x\). Khi đó $$\begin{aligned}
F(x)&=\displaystyle\int f(x)\mathrm{\,d}x=\displaystyle\int\cot x\mathrm{\,d}x\\
&=\displaystyle\int\dfrac{\cos x}{\sin x}\mathrm{\,d}x=\displaystyle\int\dfrac{1}{u}\mathrm{\,d}u\\
&=\ln|u|+C=\ln\left|\sin x\right|+C.
\end{aligned}$$
Vậy \(F(x)=\ln\left|\sin x\right|\).
Suy ra \(F\left(\dfrac{\pi}{6}\right)=\ln\left|\sin\dfrac{\pi}{6}\right|=\ln\dfrac{1}{2}=-\ln2\).