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Question

The value of 2nπ0[sinx+cosx]dx, is equal to (where [.] denotes greatest integer function)

A
nπ
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B
nπ
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C
2nπ
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D
none of these
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Solution

The correct option is A nπ

2nπ0[sinx+cosx]dx

=2nπ0[2sin(x+π4)]dx

The function is periodic with period 2π.

=2π0[2sin(x+π4)]dx

Hence, first we find the definite integration with limit from 0 to 2π

The graph of sin(x+π4) is shown in the figure. To find values of [2sin(x+π4)], we can draw the lines y=12 and y=12

2π0[2sin(x+π4)]dx

Splitting the limits for greatest integer function,

=π201.dx+3π4π20.dx+π3π41.dx+3π2π2.dx+7π43π21.dx+2π7π40.dx

=(π20)(π3π4)2(3π2π)(7π43π2)

=π

Hence, for limit 0 to 2nπ the integral will be n×π=nπ


181668_124468_ans_2c0e92b9f97542dbaff7a451ba5384da.png

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