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Evaluate int(0)^(npi+t)(|cosx|+|sinx|)dx...

Evaluate `int_(0)^(npi+t)(|cosx|+|sinx|)dx,` where `n epsilonN` and `t epsilon[0,pi//2]`.

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Verified by Experts

The correct Answer is:
`4n+sint-cost+1`
Let `I=int_(0)^(npi+1)(|cosx|+|sinx|)dx`
`=int_(0)^(npi)(|cosx|+|sinx|)dx+int_(npi)^(npi+1)(|cosx|+|sinx|)dx`
`=2nint_(0)^(pi//2) (|cosx|+|sinx|)dx+int_(0)^(1)(|cosx|+|sinx|)dx`
`=2n int_(0)^(pi//2) (cosx+sinx)dx+int_(0)^(1)(cosx+sinx)dx`
`=4n+sint-cost+1`
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