Integrate log(sinx+cosx)dx from -pi/4 to pi/4.

First put x = -x and add

\(I = int limits_{-pi/4}^{pi/4}log(sinx+cosx)dx = int limits_{-pi/4}^{pi/4}log(-sinx+cosx)dx 2I = int limits_{-pi/4}^{pi/4}log(-sin^2x+cos^2x)dx = int limits_{-pi/4}^{pi/4}log(cos2x)dx = 2int limits_{0}^{pi/4}log(cos2x)dxI =int limits_{0}^{pi/4}log(cos2x)dx \Now put 2x = t Rightarrow dx = dt/2 I =1/2int limits_{0}^{pi/2}log(cost)dt\)

Refer to the following steps to solve for the simplified integral.

\(Rightarrow intlimits_{0}^{pi/2}ln cosxdx=I=intlimits_{0}^{pi/2}ln sinxdx\\)

By symmetry we have In cos z = In sin z on the interval [0, 1/2]. This is true for any even/od

function on this interval, as is an exercise in Demidovich-Problems in Analysis.



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