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Question

Let f(x)=∣ ∣ ∣1cos2xcos2xcos2xcos2xcot2xsecxcosxsec2x+cosx.cosec2 x∣ ∣ ∣. If π/20f(x)dx=mπ+n, then the value of 4m+15n is

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Solution

f(x)=∣ ∣ ∣1cos2xcos2xcos2xcos2xcot2xsecxcosxsec2x+cosx.cosec2 x∣ ∣ ∣

R3R3secx R1
=∣ ∣ ∣ ∣1cos2xcos2xcos2xcos2xcot2x001cos2x+cosxsin2xcosx∣ ∣ ∣ ∣
=[1cos2x+cosxsin2x(1sin2x)](cos2xcos4x)
=[1cos2x+cos3xsin2x](cos2xcos4x)
=sin2x+cos5x

π/20f(x)dx=π/20sin2x+cos5x dx
=π4+815

m=14, n=8154m+15n=9

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