1.

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:-

Answer»

0
`(Pi)/(4sqrt(2))-(Pi^2)/4`
1
`cos 1 - sin 2`

Solution :Let `p(x) = sin^(2020)x.cos^(2019)x`
`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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