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π
/
2
\pi/2
π
/2
∫
0
π
/
2
(
2
log
sin
x
−
log
sin
2
x
)
d
x
equals
−
\int_{0}^{\pi/2} (2 \log \sin x - \log \sin 2x) dx \text{ equals } -
∫
0
π
/2
(
2
lo
g
sin
x
−
lo
g
sin
2
x
)
d
x
equals
−
Mathematics
Problem
π
/
2
\pi/2
π
/2
∫
0
π
/
2
(
2
log
sin
x
−
log
sin
2
x
)
d
x
equals
−
\int_{0}^{\pi/2} (2 \log \sin x - \log \sin 2x) dx \text{ equals } -
∫
0
π
/2
(
2
lo
g
sin
x
−
lo
g
sin
2
x
)
d
x
equals
−
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