Tích phân \(\int\limits^{\frac{\pi}{6}}_0\frac{\text{d}x}{\cos2x}\) bằng
\(\frac{1}{2}\ln\left(2-\sqrt{3}\right)\). \(\ln\left(2-\sqrt{3}\right)\). \(\ln\sqrt{2+\sqrt{3}}\). \(\frac{1}{3}\ln\left(2+\sqrt{3}\right)\). Hướng dẫn giải:\(I=\int\limits^{\frac{\pi}{6}}_0\frac{\text{d}x}{\cos2x}=\frac{1}{2}\int\limits^{\frac{\pi}{6}}_0\frac{2\cos2x\text{d}x}{\cos^22x}=\frac{1}{2}\int\limits^{\frac{\pi}{6}}_0\frac{2\cos2x\text{d}x}{1-\sin^22x}\)
Đặt \(t=\sin2x\) \(\Rightarrow\text{d}t=2.\cos2x\text{d}x\)
Đổi cận: \(x|^{\frac{\pi}{6}}_0\Rightarrow t|^{\frac{\sqrt{3}}{2}}_0\)
\(I=\frac{1}{2}\int\limits^{\frac{\sqrt{3}}{2}}_0\frac{\text{d}t}{1-t^2}=\frac{1}{2}\int\limits^{\frac{\sqrt{3}}{2}}_0\frac{1}{\left(1-t\right)\left(1+t\right)}\text{d}t\)
\(=\frac{1}{4}\int\limits^{\frac{\sqrt{3}}{2}}_0\left(\frac{1}{1-t}+\frac{1}{1+t}\right)\text{d}t\)
\(=\frac{1}{4}\left[-\int\limits^{\frac{\sqrt{3}}{2}}_0\frac{\text{d}\left(1-t\right)}{1-t}+\int\limits^{\frac{\sqrt{3}}{2}}_0\frac{\text{d}\left(1+t\right)}{1+t}\right]\)
\(=\frac{1}{4}\left(-\ln\left|1-t\right|+\ln\left|1+t\right|\right)|^{\frac{\sqrt{3}}{2}}_0\)
\(=\frac{1}{4}\ln\left|\frac{1+t}{1-t}\right||^{\frac{\sqrt{3}}{2}}_0\)
\(=\frac{1}{2}\ln\frac{2+\sqrt{3}}{2-\sqrt{3}}=\frac{1}{2}\ln\left(2+\sqrt{3}\right)^2=\frac{1}{2}\ln\left(2+\sqrt{3}\right)=\ln\sqrt{2+\sqrt{3}}\).
