The value of the integral ∫2π0sin2θa−bcosθdθ when a > b > 0, is
A
1
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B
π
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C
π/2
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D
0
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Solution
The correct option is D 0 Let I=∫2π0sin2θa−bcosθdθ ...(1) Using property ∫a0f(x)dx=∫a0f(a−x)dx I=∫2π0sin2(2π−θ)a−bcos(2π−θ)dθ=∫2π0sin(4π−2θ)a−bcosθdθ =−∫2π0sin2θa−bcosθdθ ...(2) Adding (1) and (2) 2I=0⇒I=0