lf I=∫a−a(αsin5x+βtan3x+γcosx)dx, where α,β,γ are constants, then the value of I depends on
A
γ,a
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B
α,β,γ,a
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C
α,β,a,
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D
β
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Solution
The correct option is Dγ,a I=∫a−a(αsin5x+βtan3x+γcosx)dx−(1) I=∫a−a[α(sin5(−x)+βtan3(−x)+γcos(−x))]dx =∫a−a[−αsin5x−βtan3x+γcosx]dx−(2) (1)+(2) 2I=∫a−aγcosxdx I=1/2∫a−aγcosxdx So, depends on a,γ