calculus
posted by alsa .
Integral of ln(sinx+cosx) with respect to x from pi/4 to pi/4

sin(x) + cos(x) = sqrt(2) sin(x + pi/4)
So, the integral can be written as:
pi/4 ln(2) + Integral of Log[sin(x)]dx from 0 to pi/2
Let's call the integral in here I:
I = Integral of Log[sin(x)]dx from 0 to pi/2
Then note that because sin is symmetric w.r.t. reflection about x = pi/2 integrating to pi will yield twice the value:
2 I = Integral of Log[sin(x)]dx from 0 to pi
Then substitute in this integral
x = 2 y:
2 I = 2 Integral of Log[sin(2y)]dy from 0 to pi/2 >
I = Integral of
(Log(2) + Log[cos(y)] +Log[sin(y)] )dy from 0 to pi/2 =
pi/2 Log(2) + 2 I
because the integral of log[cos(y)] is the same as the integral of
log[sin(y)], as cos(y) is sin(pi/2y).
Solving for I gives:
I = pi/2 log(2)
The original integral is thus
 pi/4 log(2)
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