Lời giải:
\(I=\int ^{\frac{\pi}{3}}_{0}\frac{1}{\cos ^2x}dx+\int ^{\frac{\pi}{3}}_{0}\frac{x\sin xdx}{\cos ^2x}\)
Trong đó:
\(\int ^{\frac{\pi}{3}}_{0}\frac{1}{\cos ^2x}dx=|^\frac{\pi}{3}_0\tan x=\sqrt{3}\)
\(\int ^{\frac{\pi}{3}}_{0}\frac{x\sin xdx}{\cos ^2x}=-\int ^{\frac{\pi}{3}}_{0}\frac{xd(\cos x)}{\cos ^2x}=-\int ^{\frac{\pi}{3}}_{0}xd(\frac{-1}{\cos x})\)
\(=\int ^{\frac{\pi}{3}}_{0}xd(\frac{1}{\cos x})=|^\frac{\pi}{3}_0(\frac{x}{\cos x})-\int ^{\frac{\pi}{3}}_{0}\frac{dx}{\cos x}\)
\(=\frac{2\pi}{3}-\int ^{\frac{\pi}{3}}_{0}\frac{\cos xdx}{\cos ^2x}=\frac{2\pi}{3}-\int ^{\frac{\pi}{3}}_{0}\frac{d(\sin x)}{1-\sin ^2x}\)
\(=\frac{2}{3}\pi-\frac{1}{2}\int ^{\frac{\pi}{3}}_{0}(\frac{1}{1-\sin x}+\frac{1}{1+\sin x})d(\sin x)\)
\(=\frac{2}{3}\pi +\frac{1}{2}\int ^{\frac{\pi}{3}}_{0}\frac{d(\sin x)}{\sin x-1}-\frac{1}{2}\int ^{\frac{\pi}{3}}_{0}\frac{d(\sin x)}{\sin x+1}\)
\(=\frac{2}{3}\pi +\frac{1}{2}|^\frac{\pi}{3}_0\ln |\sin x-1|-\frac{1}{2}|^\frac{\pi}{3}_0\ln |\sin x+1|\)
\(=\frac{2\pi}{3}+\ln (2-\sqrt{3})\)
Do đó $I=\sqrt{3}+\frac{2\pi}{3}+\ln (2-\sqrt{3})$
Đáp án B.
