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Today's Calculation Of Integral
2011 Today's Calculation Of Integral
673
673
Part of
2011 Today's Calculation Of Integral
Problems
(1)
Today's calculation of Integral 673
Source:
2/5/2011
Let
f
(
x
)
=
∫
0
x
1
1
+
t
2
d
t
.
f(x)=\int_0^ x \frac{1}{1+t^2}dt.
f
(
x
)
=
∫
0
x
1
+
t
2
1
d
t
.
For
−
1
≤
x
<
1
-1\leq x<1
−
1
≤
x
<
1
, find
cos
{
2
f
(
1
+
x
1
−
x
)
}
.
\cos \left\{2f\left(\sqrt{\frac{1+x}{1-x}}\right)\right\}.
cos
{
2
f
(
1
−
x
1
+
x
)
}
.
2011 Ritsumeikan University entrance exam/Science and Technology
calculus
integration
trigonometry
calculus computations