MIT Integration Bee 2009 Worked Solutions - TweetCube

MIT Integration Bee 2009 Worked Solutions
The target time is 4 minutes for eacheek!
* These ones took me longer than 10 minutes!
1. Trig identities:
ˆ π
0
ˆ
cos(3x)
2π
cos x dx =
0
= −2
= −2
= −π
ˆ π
cos 2xdx −
ˆ π
0
ˆ π
0
sin 2 xdx
0
1 − cos 2x
dx
2
sin 2x sin x
dx
cos x
2. Hyperbolic substitution (x = cosh u) & by parts:
ˆ
ˆ
ˆ
arccosh xdx = u sinh udu = u cosh u −
ˆ
= u cosh u − cosh udu
cosh udu
= u cosh u − sinh u
= xarccosh x − sinh(arccosh x)
3. Partial fractions:
ˆ 1 − x
3
ˆ
1 − x 4 dx = 1
1 − x 4 dx + 1 ˆ d(1 − x 4 )
4 1 − x
ˆ
4
1
=
(1 − x 2 )(1 + x 2 ) dx + 1 4 log(1 − x4 )
= 1 ˆ
1
2 1 + x 2 + 1
1 − x 2 dx + 1 4 log(1 − x4 )
= 1 2 arctan x + 1 ˆ
1
4 1 + x + 1
1 − x dx + 1 4 log(1 − x4 )
= 1 2 arctan x + 1 [
log(1 + x) − log(1 − x) + log(1 − x 4 ) ]
4
= 1 2 arctan x + 1 [ [
log(1 + x) + log (1 − x 2 )(1 + x 2 )/(1 − x) ]]
4
= 1 2 arctan x + 1 [
2 log(1 + x) + log(1 + x 2 ) ]
4
4. Partial fractions:
ˆ 2 − x
3
x 4 − x dx = ˆ
= −
1 − x 3 ˆ
x(x 3 − 1) dx + ˆ ˆ 1
x dx + ˆ 1
= − log x +
= −2 log x + 1 3
1
x(x 3 − 1) dx
1
x(x − 1)(x 2 + x + 1) dx
2x + 1
3 x 2 + x + 1 + 1
3(x − 1) − 1 x dx
ˆ d(x 2 + x + 1)
x 2 + x + 1 + 1 log(x − 1)
3
= −2 log x + 1 3 log [ (x 2 + x + 1)(x − 1) ]
= −2 log x + 1 3 log(x3 − 1)
1

5. * Substitution of t = π − x (dt = −dx) which leads to sin x = cos t so arccos(sin x) = t (I realised it
2
was this simple only after my rst thought, wanting to substitute sin x = cos t and then doing all those
motions, which worked ne except for an ambiguity in sign and an awkward limit):
ˆ 1
0
arccos(sin x)dx = −
=
ˆ π/2−1
π/2
ˆ π/2
π/2−1
] π/2
tdt
[ t
2
=
2
π/2−1
= (π − 1)/2
This one is deceptively simple. I was sure it was wrong until I did a numerical integration to check!
tdt
6. Trig identities:
ˆ π/2
−π/2
√
1 − cos xdx = 2
ˆ π/2
= 2 √ 2
= −4 √ 2
0
ˆ π/2
0
= 4( √ 2 − 1)
√
1 − (1 − 2 sin 2 x 2 )dx
( x
)
sin dx
2
[ ( x
)] π/2
cos
2
0
7. * Substitution (y = 7 − x) & trickery (someone likes the number 7 in this question, even the limits
3 + 4 = 7!):
I =
ˆ 4
3
x 1/7
ˆ 3
(7 − x) 1/7 + x dx = − (7 − y) 1/7
dy
1/7 y 1/7 + (7 − y)
1/7
=
=
4
ˆ 4
3
ˆ 4
3
= 1 − I
⇐⇒ I = 1/2
(7 − y) 1/7 + y 1/7 − y 1/7
(7 − y) 1/7 + y 1/7 dy
1 −
y 1/7
dy
(7 − y) 1/7 + y1/7 [My thanks to Mark for that! I didn't spot the add/subtract trick and was baed.]
8. Trig substitution (x = 2 sin u) & identities:
ˆ 2
1
√
4 − x2 dx = 2
ˆ π/2
= 4
= 2
π/6
ˆ π/2
π/6
ˆ π/2
π/6
√
4 − 4 sin 2 u cos udu
cos 2 udu
cos 2u + 1du
= [sin 2u + 2] π/2
π/6
= 2π 3 − √
3
2
2

9. * Hyperbolic substitution (x = sinh u, then w = 1/ sinh u, then z = w 2 + 1; i.e. z = coth 2 u) &
identities:
ˆ
ˆ
1
x 2√ x 2 + 1 dx = cosh u
sinh 2 u
√sinh 2 u + 1 du
ˆ
1
=
sinh 2 u du
ˆ
−w
= √
w2 + 1 dw
N.B. If you recall ´
= − 1 2
ˆ
= − √ z
= − coth u
1
√ z
dz
= − coth(arcsinh x)
1
sinh 2 u du = ´ cosech 2 u du = − coth u then you're done easily. Unlike me :(
10. Trig identities, hyperbolics & substitution (u = cot x, then u = sinh t):
ˆ
ˆ
2 csc 3 xdx = csc x csc 2 xdx
=
ˆ
− csc x d cot x
dx
dx
ˆ √
= − 1 + cot 2 xd cot x
ˆ √1
= − + u2 du
ˆ
= − cosh 2 t dt
= − 1 ˆ
cosh 2t + 1 dt
2
= − (sinh 2t + 2t) /4
= − (sinh t cosh t + t) /2
[
= − u √ ]
1 + u 2 + arcsinh u /2
= − [cot x csc x + arcsinh(cot x)] /2
* This took under 10 minutes to get the idea out, but it took nearly 20 minutes for me to convince
myself it is right! Not a terribly pleasant solution.
11. Trig substitution (x = sin θ, then u = θ + π/4) & identities:
ˆ 2
0
ˆ
4
0
x + √ 4 − x dx = 4 2
= 4
π/2
ˆ π/2
0
= 4 √
2
ˆ π/2
= 2 √ 2
= 2 √ 2
−2 sin θ
2 cos θ + 2 sin θ dθ
sin θ
cos θ + sin θ dθ
0
ˆ 3π/4
π/4
ˆ 3π/4
π/4
sin θ
sin(θ + π/4) dθ
sin(u − π/4)
du
sin u
( π
)
cos − sin
4
= 2 [u − log(sin u)] 3π/4
π/4
= π
( π
4
)
cot u du
3

12. By parts (twice):
ˆ
I =
ˆ
e −x cos xdx = −e −x cos x − e −x sin xdx
ˆ
= −e −x cos x + e −x sin x − e −x cos xdx
= e −x (sin x − cos x) − I
⇐⇒ I = e −x (sin x − cos x)/2
4