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Question

The value of 02π1+sinx2dx is
(a) 0
(b) 2
(c) 8
(d) 4

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Solution

(c) 8

02π1+sinx2dx=02πsin2x4+cos2x4+2sinx4cosx4 dx=02πsinx4+cosx4dx=-cosx414+sinx41402π=4sinx4-cosx402π=4sin2π4-cos2π4-sin 0+cos 0=4sinπ2-cosπ2-0+1=41-0-0+1=4×2=8

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