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Today's Calculation Of Integral
2005 Today's Calculation Of Integral
39
39
Part of
2005 Today's Calculation Of Integral
Problems
(1)
Today's calculation of Integral 39
Source: 1990 Tokyo Institute of Technology
6/9/2005
Find the minimum value of the following function
f
(
x
)
f(x)
f
(
x
)
defined at
0
<
x
<
π
2
0<x<\frac{\pi}{2}
0
<
x
<
2
π
.
f
(
x
)
=
∫
0
x
d
θ
cos
θ
+
∫
x
π
2
d
θ
sin
θ
f(x)=\int_0^x \frac{d\theta}{\cos \theta}+\int_x^{\frac{\pi}{2}} \frac{d\theta}{\sin \theta}
f
(
x
)
=
∫
0
x
cos
θ
d
θ
+
∫
x
2
π
sin
θ
d
θ
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