Tích phân \(\int\limits^{\dfrac{\pi}{3}}_{\dfrac{\pi}{6}}\dfrac{\ln\left(\sin x\right)}{\cos^2x}\text{d}x\) bằng
\(\dfrac{2}{\sqrt{3}}\ln\dfrac{3}{4}-\dfrac{\pi}{6}\).\(\dfrac{1}{\sqrt{3}}\ln\dfrac{3\sqrt{3}}{4}-\dfrac{\pi}{6}\).\(\dfrac{\sqrt{3}}{2}\ln\dfrac{3}{4}+\dfrac{\pi}{6}\).\(\dfrac{2}{\sqrt{3}}\ln\dfrac{3\sqrt{3}}{4}+\dfrac{\pi}{6}\).Hướng dẫn giải:Biến đổi tích phân \(\int\limits\dfrac{\ln\left(\sin x\right)}{\cos^2x}\text{d}x=\int\ln\left(\sin x\right)\text{d}\left(\tan x\right)=\ln\left(\sin x\right)\left(\tan x\right)-\int\limits\tan x\text{d}\left(\ln\left(\sin x\right)\right)=\)\(\left(\tan x\right)\ln\left(\sin x\right)-\int\tan x.\dfrac{\cos x}{\sin x}\text{d}x=\left(\tan x\right)\ln\left(\sin x\right)-\int\text{d}x=\left(\tan x\right)\ln\left(\sin x\right)-x+C.\)
Do đó
\(\int\limits^{\frac{\pi}{3}}_{\frac{\pi}{6}}\dfrac{\ln\left(\sin x\right)}{\cos^2x}\text{d}x=\left(\tan x\right)\ln\left|\sin x\right||^{\frac{\pi}{3}}_{\frac{\pi}{6}}-x|^{\frac{\pi}{3}}_{\frac{\pi}{6}}=\left(\sqrt{3}\ln\dfrac{\sqrt{3}}{2}-\dfrac{1}{\sqrt{3}}\ln\dfrac{1}{2}\right)-\dfrac{\pi}{6}=\sqrt{3}\left(\ln\sqrt{3}-\ln2\right)+\dfrac{1}{\sqrt{3}}\ln2-\dfrac{\pi}{6}\)
\(=\dfrac{1}{\sqrt{3}}\left(3\ln\sqrt{3}-2\ln2\right)-\dfrac{\pi}{6}=\dfrac{1}{\sqrt{3}}\ln\dfrac{3\sqrt{3}}{4}-\dfrac{\pi}{6}.\)