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If Δ(x)=1cosx1-cosx1+sinxcosx1+sinx-cosxsinxsinx1 then ∫0π/2Δ(x)dx equal

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a

14

b

12

c

0

d

-12

answer is D.

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Detailed Solution

Using , C1→C1-C2-C3, we get Δ(x)=0cosx1-cosx0cosx1+sinx-cosx-1sinx1=(-1)cosx1-cosxcosx1+sinx-cosx=(-1)cosx[1+sinx-cosx-1+cosx]=-cosxsinx=-12(sin2x)∴ ∫0π/2Δ(x)dx=-12(-1)2cos2x0π/2=14(cosπ-cos0)=-12

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If Δ(x)=1cosx1-cosx1+sinxcosx1+sinx-cosxsinxsinx1 then ∫0π/2Δ(x)dx equal