JEE Main 2002Annual, Paper 1MathematicsMediumIntegral Calculus369× since 2002∫−ππ2x(1+sinx)1+cos2xdx\int_{ - \pi }^\pi {{{2x\left( {1 + \sin x} \right)} \over {1 + {{\cos }^2}x}}} dx∫−ππ1+cos2x2x(1+sinx)dx isAOption A. π24{{{\pi ^2}} \over 4}4π2BOption B. π2{{\pi ^2}}π2COption C. zeroDOption D. π2{\pi \over 2}2πCheckSkip · show solution
