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Mathematics
Mathematics
Problem
Prove that
∫
0
π
/
2
log
(
sin
x
)
d
x
=
∫
0
π
/
2
log
(
cos
x
)
d
x
=
−
π
2
log
2
\int_{0}^{\pi/2} \log (\sin x) dx = \int_{0}^{\pi/2} \log (\cos x) dx = -\frac{\pi}{2} \log 2
∫
0
π
/2
lo
g
(
sin
x
)
d
x
=
∫
0
π
/2
lo
g
(
cos
x
)
d
x
=
−
2
π
lo
g
2
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