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Today's Calculation Of Integral
2005 Today's Calculation Of Integral
21
Today's calculation of Integral 21
Today's calculation of Integral 21
Source: From Entrance Examination for Japanese University 2004
May 21, 2005
calculus
integration
logarithms
trigonometry
calculus computations
Problem Statement
[1] Tokyo Univ. of Science:
∫
ln
x
(
x
+
1
)
2
d
x
\int \frac{\ln x}{(x+1)^2}dx
∫
(
x
+
1
)
2
l
n
x
d
x
[2] Saitama Univ.:
∫
5
3
sin
x
+
4
cos
x
d
x
\int \frac{5}{3\sin x+4\cos x}dx
∫
3
s
i
n
x
+
4
c
o
s
x
5
d
x
[3] Yokohama City Univ.:
∫
1
3
1
x
2
+
1
d
x
\int_1^{\sqrt{3}} \frac{1}{\sqrt{x^2+1}}dx
∫
1
3
x
2
+
1
1
d
x
[4] Daido Institute of Technology:
∫
0
π
2
sin
3
x
sin
x
+
cos
x
d
x
\int_0^{\frac{\pi}{2}} \frac{\sin ^ 3 x}{\sin x +\cos x}dx
∫
0
2
π
s
i
n
x
+
c
o
s
x
s
i
n
3
x
d
x
[5] Gunma Univ.:
∫
0
3
π
4
{
(
1
+
x
)
sin
x
+
(
1
−
x
)
cos
x
}
d
x
\int_0^{\frac{3\pi}{4}} \{(1+x)\sin x+(1-x)\cos x\}dx
∫
0
4
3
π
{(
1
+
x
)
sin
x
+
(
1
−
x
)
cos
x
}
d
x
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