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Today's Calculation Of Integral
2005 Today's Calculation Of Integral
60
60
Part of
2005 Today's Calculation Of Integral
Problems
(1)
Today's Calculation of Integral 60
Source: 1978 Hokkaido University
7/1/2005
Let
a
n
=
∫
0
π
2
sin
2
t
(
1
−
sin
t
)
n
−
1
2
d
t
(
n
=
1
,
2
,
⋯
)
a_n=\int_0^{\frac{\pi}{2}} \sin 2t\ (1-\sin t)^{\frac{n-1}{2}}dt\ (n=1,2,\cdots)
a
n
=
∫
0
2
π
sin
2
t
(
1
−
sin
t
)
2
n
−
1
d
t
(
n
=
1
,
2
,
⋯
)
Evaluate
∑
n
=
1
∞
(
n
+
1
)
(
a
n
−
a
n
+
1
)
\sum_{n=1}^{\infty} (n+1)(a_n-a_{n+1})
n
=
1
∑
∞
(
n
+
1
)
(
a
n
−
a
n
+
1
)
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