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SOMEVERYCHALLENGINGCALCULUSPROBLEMS
Joseph Breen
Here are two difficult calculus problems, solved using only (sophisticated and clever applications of)
elementary calculus. In particular, there is no complex analysis or use of the residue theorem, Fourier
series, or anything like that. Both problems were the basis for talks given at the UCLA GSO Seminar.
The integral is the concatenation of two integrals from [3]. The infinite series was originally evaluated
byothermethodsin[2],andthesolutionpresentedbelowisinspiredbythesolutionfrom[4],togetherwith
other computations found on the internet and my own computational decisions.
Contents
1 AReallyHardIntegral 1
2 AReallyHardInfiniteSeries 6
1 AReallyHardIntegral
Theintegral we will evaluate is
Z π
I = 2 arccos cosx dx. (1)
0 1+2cosx
Step 1: Rewrite the integrand with trigonometry and then introduce a double integral.
2
Webeginwithsometrigonometry. Recall the double angle identity cos(2θ) = 2cos θ −1. This implies
q
2 2 1+α
2θ = arccos 2cos θ −1 . Letting α = 2cos θ −1 yields θ = arccos 2 , and thus
!
arccos(α) = 2arccos r1+α .
2
Usingthis, the integral becomes
!
Z π r
2 1+3cosx
I = 2arccos 2+4cosx dx.
0 | {z }
θ∗
Next, consider a right triangle with angle θ∗.
1 q1+cosx
2+4cosx
θ∗
q1+3cosx
2+4cosx
This picture implies ! !
arccos r1+3cosx =arctan r 1+cosx
2+4cosx 1+3cosx
1
andso !
Z π r
2 1+cosx
I = 2 arctan 1+3cosx dx.
0
With the goal of using the aforementioned double angle identity again, we make the substitution x = 2y.
Thendx=2dyandwehave
Z π r
4 1+cos2y
I = 4 arctan 1+3cos2y dy
0 s !
Z π 2
4 2cos y
=4 arctan −2+6cos2y dy
0
Z π !
4 cosy
=4 0 arctan p 2 dy.
2−3sin y
| {z }
b
Recall that R 21 2 dx = 1 arctan(x)+C. Thus,
a +x a a
Z 1 1 1 Z 1 1 1 1
dt = dt = arctan(bt)
0 1+b2t2 b2 0 b−2 +t2 b 0
= 1 arctan(b).
b
This implies
Z π Z 1
4 cosy 1
I = 4 p 2 dtdy
0 2 0 1+ cos y t2
2−3sin y 2−3sin2y
Z π Z 1 p 2
4 cosy 2−3sin y
=4 2 2 2 dtdy
0 0 2−3sin y+cos yt
Z π Z 1 p 2
4 cosy 2−3sin y
=4 2 2 2 dtdy.
0 0 (t +2)−(t +3)sin y
Next, we adjust constants in order to simplify the expression in the numerator. In particular, let siny =
q2sinw. Thencosydy =q2coswdw,andtheintegralbecomes
3 3
Z π Z 1 √ r
3 2cosw 2
I = 4 2 2 2 2 dt 3 coswdw
0 0 (t +2)−(t +3)3sin w
√ Z π Z 1 2
3 cos w
=8 3 2 2 2 dtdw
0π 0 (3t +6)−(2t +6)sin w
√ Z 3 Z 1 cos2w
=8 3 2 2 2 dtdw
0 0 t +(2t +6)cos w
√ Z π Z 1
3 1
=8 3 t2 sec2 w + (2t2 + 6) dtdw.
0 0
Step 2: Use a trig substitution, partial fractions, then integration by parts.
2 2 2 2 2 1
Lets = tanw. Thends = sec wdw. Since1+tan w = sec w,wehavesec w = 1+s anddw = 1+s2 ds.
Thus,
Z √ Z
√ 3 1 1 1
I = 8 3 2 2 2 dt 2 ds
0 0 (1+s )t +(2t +6) 1+s
Z √ Z
√ 3 1 1
=8 3 (1+s2)(3t2 +t2s2 +6) dtds.
0 0
2
Next, we decomposetheintegrandwithpartialfractions. We have
√ Z √3Z 1 1 1 t2
I = 8 3 0 0 2t2 +6 1+s2 − 3t2 +t2s2 +6 dtds.
Thetermsintheparenthesescanbeintegratedwithrespecttosusingtheinversetangent. Thus,weswitch
the order of integration.
√ Z 1Z √3 1 1 1
I = 8 3 2 2 − 6 2 dsdt
0 0 2t +6 1+s 3+t2 +s
√ Z 1 1 π 1 √3
=4 3 0 t2+33 −q 6 arctanq 6 dt.
3+ 2 3+ 2
t t
HerewehaveagainusedthefactthatR 21 2 dx = 1 arctan(x)+C. Next,the 21 termcanbeintegrated
a +x a a t +3
similarly with respect to t. This gives
√ Z 1 √ Z 1 √
I = 4π 3 1 dt −4 3 1 q 1 arctanq 3 dt
3 0 t2 +3 0 (t2 +3) 3+ 6 3+ 6
2 2
√ t t
4π 3 1 1 Z 1 1 t t
= 3 √ arctan √ −4 (t2 + 3) √ 2 arctan √2 dt
3 3 0 t +2 t +2
2π2 Z 1 t t
= 9 −4 2 √2 arctan √2 dt.
0 (t +3) t +2 t +2
Next, we will integrate by parts. Let
t t
u=arctan √ 2 and dv = 2 √2 dt.
t +2 (t +3) t +2
Then √
2 t2
1 t +2−√2 1
du = · t +2 dt = √ dt. (2)
1+ t2 t2 +2 (t2 + 1) t2 +2
2
t +2
Next, observe that
d p2 1 t t
dx arctan( t +2)= (1+t2+2 · √ 2 = 2 √2 .
t +2 (t +3) t +2
√2
Thus, v = arctan( t +2). Integrating by parts with this set up yields
√
1 Z
2π2 t p 1 arctan( t2 +2)
I = −4arctan √ arctan( t2 +2) − √ dt
2 2 2
9 t +2 0 (t +1) t +2
! 0
2 Z 1 √2
2π π π arctan( t +2)
= 9 −4 6 · 3 − 2 √2 dt
0 (t +1) t +2
Z 1 √2
arctan( t +2)
=4 2 √2 dt.
0 (t +1) t +2
Step 3: Differentiate under the integral.
Weintroduceanadditionalparameterintheintegrand:
Z 1 arctan(z√t2 +2)
I(z) = 4 2 √2 dt.
0 (t +1) t +2
3
WeseekI =I(1). Bythefundamentaltheoremofcalculus,
Z ∞I′(z)dz = lim I(z)−I(1) ⇒ I = lim I(z)−Z ∞I′(z)dz.
1 z→∞ z→∞ 1
Wecomputeeachtermseparately.
Z 1 √2
lim arctan(z t +2)
lim I(z) = 4 z→∞ √ dt
z→∞ 2 2
0 (t +1) t +2
Z 1 π
2
=4 2 √2 dt
0 (t +1) t +2
1
t
=2πarctan √
2
t +2
0
π2
= 3 .
Thesecondtolastequalitycomesfromourpreviouscomputationin(2).
Next,
Z ∞ Z ∞Z 1 √2 !
′ d arctan(z t +2)
I (z)dz = 4 dz 2 √2 dtdz
1 1 0 (t +1) t +2
Z ∞Z 1 1 1 p2
=4 2 √2 · 1 + z2(t2 + 2) · t +2dtdz
1 0 (t +1) t +2
=4Z ∞Z 1 2 1 2 2 2 dtdz.
1 0 (t +1)(1+z t +2z )
Wedecomposetheintegrandwithpartialfractions.
Z ∞I′(z)dz = 4Z ∞Z 1 1 1 − z2 dtdz
1 1 0 1+z2 1+t2 1+z2t2+2z2
=4Z ∞ 1 Z 1 1 dt −Z 1 z2 dt dz
1 1+z2 0 1+t2 0 1+z2t2+2z2
=4Z ∞ 1 π −Z 1 1 dt dz
1+z2 4 2+ 1 +t2
1 0 z2
Z ∞
∞ 1 1 1
=πarctan(z) −4 · q arctanq dz
1+z2
1 1 2+ 1 2+ 1
z2 z2
π2 Z ∞ 1 1 1
= 4 −4 1 1+z2 · q 1 arctanq 1 dz.
2+z2 2+z2
Next, let r = 1. Then dz = − 1 dr, and the above integral becomes
z r2
Z ∞I′(z)dz = π2 −4Z 1 1 1 · √ 1 arctan√ 1 1 dr
4 1+ 2+r2 2+r2 r2
1 0 r2
π2 Z 1 1 1
= 4 −4 2 √ 2 arctan √ 2 dr.
0 (r +1) 2+r 2+r
Recall that Z √
1 arctan( t2 +2)
I = 4 2 √2 dt.
0 (t +1) t +2
4
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