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Evaluate int(0)^(pi//2)|sinx-cosx|dx....

Evaluate `int_(0)^(pi//2)|sinx-cosx|dx`.

Text Solution

Verified by Experts

The correct Answer is:
`2sqrt(2)-1)`
`int_(0)^(pi//2)|sinx-cosx|dx`
`=int_(0)^(pi//4)-(sinx-cosx)dx+int_(pi//4)^(pi//2)(sinx-cosx)dx`
`=|cosx+sinx|_(0)^(pi//4)+|-cosx-sinx|_(pi//4)^(pi//2)`
`=(1/(sqrt(2))+1/(sqrt(2))-1-0)+(-0-1+1/(sqrt(2))+1/(sqrt(2)))`
`=4/(sqrt(2))-2=2sqrt(2)-2`
`2(sqrt(2)-1)`
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Knowledge Check

  • int_(0)^(pi//2) abs(sinx-cosx) dx=

    A
    0
    B
    `sqrt(2) -1`
    C
    `2sqrt(2)-2`
    D
    `2sqrt(2) +2`

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