MathematicsHard210×since 2002Q3460The value of e−π4+∫0π4e−xtan50xdx∫0π4e−x(tan49x+tan51x)dx{{{e^{ - {\pi \over 4}}} + \int\limits_0^{{\pi \over 4}} {{e^{ - x}}{{\tan }^{50}}xdx} } \over {\int\limits_0^{{\pi \over 4}} {{e^{ - x}}({{\tan }^{49}}x + {{\tan }^{51}}x)dx} }}0∫4πe−x(tan49x+tan51x)dxe−4π+0∫4πe−xtan50xdx isA51B50C25D49Check answerSkip