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If int_(0)^(2Pi) sin^(2020) x.cos^(2019) x dx = 1 then the value of f(I), (where f(x) = x cos(I) – x^(2) sin2(I)) is equal to:- |
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Answer» 0 `implies p(2pi-x)=p(x)` `implies I = 2.int_(0)^(pi) sin^(2020)(pi-x).cos^(2019)(pi-x) dx` `implies I = -2int_(0)^(pi)sin^(2020)x cos^(2019)x dx` `implies 3I = 0 or I = 0 implies f (I) = 0`. |
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