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Computational Essay

PeterCullenBurbery
In this essay I will solve problems from the 2022 MIT Integration Bee Regular Season Problems.
In[]:=
100
∫
0

x
x
Out[]=
715
In[]:=

Log[1+x]
2
x
x
Out[]=
Log[x]-Log[1+x]-
Log[1+x]
x
In[]:=

Log[1+x]
2
x
x//FullSimplify//TraditionalForm
Out[]//TraditionalForm=
log(x)-
(x+1)log(x+1)
x
In[]:=
π
2
+1
∫
π
2
-1
Cos[ArcSin[ArcCos[Sin[x]]]]x
Out[]=
π
2
In[]:=
2
∫
-2
Abs[(x-2)(x-1)x(x+1)(x+2)]x
Out[]=
19
3
In[]:=
TimeConstrained∫(2020
2019
Sin[x]
2019
Cos[x]
-8084
2021
Sin[x]
2021
Cos[x]
)x,8
Out[]=
$Aborted
I gave up and decided to solve the integral symbolically.
In[]:=
∫b
b-1
Sin[x]
b-1
Cos[x]
-8084
c
Sin[x]
c
Cos[x]
x
Out[]=
-2+b
Cos[x]
2-b
2
(
2
Cos[x]
)
Hypergeometric2F1
2-b
2
,
b
2
,
2+b
2
,
2
Sin[x]

b
Sin[x]
-
1
1+c
8084
-1+c
Cos[x]
1-c
2
(
2
Cos[x]
)
Hypergeometric2F1
1-c
2
,
1+c
2
,
3+c
2
,
2
Sin[x]

1+c
Sin[x]
In[]:=
∫b
b-1
Sin[x]
b-1
Cos[x]
-4c
c
Sin[x]
c
Cos[x]
x
Out[]=
-2+b
Cos[x]
2-b
2
(
2
Cos[x]
)
Hypergeometric2F1
2-b
2
,
b
2
,
2+b
2
,
2
Sin[x]

b
Sin[x]
-
1
1+c
4c
-1+c
Cos[x]
1-c
2
(
2
Cos[x]
)
Hypergeometric2F1
1-c
2
,
1+c
2
,
3+c
2
,
2
Sin[x]

1+c
Sin[x]
In[]:=
N∫b
b-1
Sin[x]
b-1
Cos[x]
-4c
c
Sin[x]
c
Cos[x]
x/.{b->2020.,c->2021.}
Out[]=
2018.
Cos[x]
2020.
Sin[x]
(
⋯1⋯
)
1009.
(
2
Cos[x]
)
-
3.99802
⋯2⋯
(
⋯1515⋯
+0.500247
⋯1⋯
)
1010.
(
2
Cos[x]
)
Full expression not available
(
original memory size:
405.2 kB)
I think Mathematica loses this round.
In[]:=

3
3
x
+2
2
x
+1
3
3
x
+1
x
Out[]=
(1+x)
2/3
(1+
3
x
)
In[]:=

3
3
x
+2
2
x
+1
3
3
x
+1
x//FullSimplify//TraditionalForm//AbsoluteTiming
Out[]=
10.031,(x+1)
2/3
(
3
x
+1)

In[]:=

1
4
Sin[x]
4
Cos[x]
x
Out[]=
-
8Cot[x]
3
-
1
3
Cot[x]
2
Csc[x]
+
8Tan[x]
3
+
1
3
2
Sec[x]
Tan[x]
In[]:=

1
4
Sin[x]
4
Cos[x]
x//FullSimplify//TraditionalForm//AbsoluteTiming
Out[]=
0.0696823,
1
6
(cos(6x)-3cos(2x))
3
csc
(x)
3
sec
(x)
Mathematica loses this round in terms of a simpler solution.
In[]:=

x+Sin[x]
1+Cos[x]
x//TraditionalForm
Out[]//TraditionalForm=
xtan
x
2
In[]:=
∫
3
Sinh[x]
2
Cosh[x]
xx//TraditionalForm
Out[]//TraditionalForm=
450sinh(x)+25sinh(3x)-9sinh(5x)-450xcosh(x)-75xcosh(3x)+45xcosh(5x)
3600
In[]:=
∫
3
Sinh[x]
2
Cosh[x]
xx//FullSimplify//TraditionalForm
Out[]//TraditionalForm=
450sinh(x)+25sinh(3x)-9sinh(5x)+120x(3cosh(2x)-7)
3
cosh
(x)
3600
Mathematica also loses this round to the MIT bee unfortunately.
∫
expr

var
In[]:=
∫
x
4
x
2
3
x//FullSimplify//TraditionalForm
Out[]//TraditionalForm=
x
2
3
(
x
2
log(3)-1)
log(2)
2
log
(3)
I think Mathematica wins this because Mathematica's answer is slightly simpler.
I think Mathematica has a more complex form than the answer.
This outputted an error message but Mathematica still got the right answer!
I wasn't able to obtain a solution for this.
I'm not sure how to solve this problem: