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Today's Calculation Of Integral
2010 Today's Calculation Of Integral
666
666
Part of
2010 Today's Calculation Of Integral
Problems
(1)
Today's calculation of Integral 666
Source:
1/12/2011
Let
f
(
x
)
f(x)
f
(
x
)
be a function defined in
0
<
x
<
π
2
0<x<\frac{\pi}{2}
0
<
x
<
2
π
satisfying:(i)
f
(
π
6
)
=
0
f\left(\frac{\pi}{6}\right)=0
f
(
6
π
)
=
0
(ii)
f
′
(
x
)
tan
x
=
∫
π
6
x
2
cos
t
sin
t
d
t
f'(x)\tan x=\int_{\frac{\pi}{6}}^x \frac{2\cos t}{\sin t}dt
f
′
(
x
)
tan
x
=
∫
6
π
x
s
i
n
t
2
c
o
s
t
d
t
.Find
f
(
x
)
f(x)
f
(
x
)
.1987 Sapporo Medical University entrance exam
calculus
integration
function
trigonometry
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