MathematicsMedium64×since 2004Q4000If the ∫5tanxtanx−2dx=x+a ln ∣sinx−2cosx∣+k,\int {{{5\tan x} \over {\tan x - 2}}dx = x + a\,\ln \,\left| {\sin x - 2\cos x} \right| + k,}∫tanx−25tanxdx=x+aln∣sinx−2cosx∣+k, then aaa is equal to :A−1-1−1B−2-2−2C111D222Check answerSkip