∫10π+π610π+π3 (sin⁡x+cos⁡x)dx is equal to

10π+π610π+π3(sinx+cosx)dx is equal to

  1. A

    3

  2. B

    1+3

  3. C

    3-1

  4. D

    1

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    Solution:

    f(x)=sinx+cosx is periodic with period 2πlet,  I=10π+π/610π+π/3(sinx+cosx)dx=π/6π/3(sinx+cosx)dx=sinx-cosxπ3π6=32121232=(31)

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