Find the integral of log(sinx+cosx) from -pi/4 to pi/4.

\(I = int_{-pi/4}^{pi/4}log(sinx+cosx)dx\\) \(I = int_{-pi/4}^{pi/4}log[sqrt{2}]*[frac{1}{sqrt{2}}sinx+frac{1}{sqrt{2}}cosx]=int_{-pi/4}^{pi/4}log(sqrt{2}sin[x+pi/4])dx\)

Put x+π/4=t

\(I = int_{0}^{pi/2}log[sqrt{2}]sintdt\=int_{0}^{pi/2}log(sqrt{2}+log sintdt\=logsqrt{2}int_{0}^{pi/2}1dt+int_{0}^{pi/2}log sintdt\=pi/2 logsqrt{2}-pi/2 log2\=-pi/4 log2\)

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