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Question

π/201sinx+cosxdx

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Solution

π201(sinx+cosx)dxπ2012sin(x+π4)dx=π2012cosec(x+π4)12[logcosec(x+π4)cot(x+π4)]+c12[logtan(x2+π8)]π20=12[log0.240.41].

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