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−
1
/
2
1
/
2
(
cos
x
)
[
log
(
1
−
x
1
+
x
)
]
d
x
=
\int_{-1/2}^{1/2} (\cos x) \left[ \log \left( \frac{1-x}{1+x} \right) \right] dx =
∫
−
1/2
1/2
(
cos
x
)
[
lo
g
(
1
+
x
1
−
x
)
]
d
x
=
Mathematics
Problem
∫
−
1
/
2
1
/
2
(
cos
x
)
[
log
(
1
−
x
1
+
x
)
]
d
x
=
\int_{-1/2}^{1/2} (\cos x) \left[ \log \left( \frac{1-x}{1+x} \right) \right] dx =
∫
−
1/2
1/2
(
cos
x
)
[
lo
g
(
1
+
x
1
−
x
)
]
d
x
=
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