QuestionReportMathematics Question on Definite Integral∫π6π311+cotxdx=\int\limits_\frac{\pi}{6}^\frac{\pi}{3}\frac{1}{1+\sqrt{cotx}}dx=6π∫3π1+cotx1dx=Aπ12\frac{\pi}{12}12πBπ212\frac{\pi^2}{12}12π2Cπ6\frac{\pi}{6}6πDπ2\frac{\pi}{2}2πAnswerπ12\frac{\pi}{12}12πExplanationSolutionThe correct option is (A):π12\frac{\pi}{12}12π