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Question

If π/40(cos2θ)32cosθdθ=aπ16b, then the value of a+b is:
(where a and b are relative prime)

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Solution

I=π/40(12sin2θ)32cosθdθLetsinα=2sinθcosαdα=2cosθdθI=π/20(cos2α)3212cosαdαI=12π/20cos4αdαπ/20cosnxdx=(n1)n(n3)(n2)12π2, when n is evenI=12×34×12×π2=3π162a=3,b=2a+b=5

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