Solutions by Mike Hirschhorn

The 2014 Integration Bee—Solutions and comments
Mike Hirschhorn
Qualifying Round
1.
Z
2
3
Z
4
3x (x − 1) dx =
u4 du =
1
1 5
u + C = (x3 − 1)5 + C.
5
5
2.
Z
√
√
5
2
√
hp
i√5
2
dx =
x − 1 √ = 2 − 1 = 1.
x
x2 − 1
2
3.
Z
sin
2
x
2
dx =
Z
1 1
1
1
− cos x dx = x − sin x + C.
2 2
2
2
4.
Z
1
1
−1 x
+ C.
dx
=
tan
a2 + x 2
a
a
5.
Z
3x − 2
dx =
2x + 1
Z
3 7 1
3
7
−
dx = x − log |2x + 1| + C.
2 2 2x + 1
2
4
6.
Z
π
2
0
1
dθ =
1 + sin θ
Z
=
Z
1
0
1
1+
1
0
2t
1 + t2
·
2
dt
1 + t2
h
2
2 i1
= −1 + 2 = 1.
dt
=
−
(1 + t)2
1+t 0
7.
Z
x2014 dx =
x2015
+ C.
2015
8.
Z
1
dx =
x2 − 4
Z
1
4
1
1
−
x−2 x+2
x − 2
1
dx = log + C.
4
x+2
Typeset by AMS-TEX
1
2
9.
Z
ex (cos x + sin x) dx = Aex cos x + Bex sin x + C
for some A, B. Differentiating with respect to x gives
ex (cos x + sin x) = (A + B)ex cos x + (−A + B)ex sin x.
Comparing coefficients gives A + B = 1, −A + B = 1, A = 0, B = 1, so
Z
ex (cos x + sin x) dx = ex sin x + C.
Alternatively, you can do two integrations by parts.
10.
1
dx =
x log x
Z
Z
1
du = log |u| + C = log | log x| + C.
u
11.
Z
π x dx =
Z
ex log π dx =
1
1 x log π
e
+C =
π x + C.
log π
log π
12.
Z
1
1
1
1
dx = · − + C = −
+ C.
2
3x
3
x
3x
13.
Z
2
2x
1
p
x2 − 1 dx =
h2
3
3
(x2 − 1) 2
i2
1
=
√
2 3
(3 2 − 0) = 2 3.
3
14.
Z
x2 + 3
x2 + 3
dz
=
z + C.
x2 − 3
x2 − 3
(We must assume x is a constant function of z.)
15.
Z
x5 log |x| dx =
x6
log |x| −
6
Z
x5
x6
x6
dx =
log |x| −
+C
6
6
36
by integration by parts with u = log |x|, dv = x5 dx.
3
16.
Let
Z
π
0
x sin x
dx = I. If we make the substitution x = π − u, we find
1 + cos2 x
I=
Z
π
0
(π − u) sin u
du.
1 + cos2 u
It follows that
2I =
Z
π sin x
dx = π
1 + cos2 x
Z
1
−1
h
i1
π π π 2
1
−1
=
− −
,
du
=
π
tan
u
=
π
1 + u2
4
4
2
−1
so
Z
π
0
π2
x sin x
dx
=
I
=
.
1 + cos2 x
4
17.
√
Z
1 + sin 2x dx =
Z p
(cos x +
sin x)2
dx =
Z
cos x+sin x dx = − cos x+sin x+C.
√
(Not exactly legitimate, I’m afraid. It is not always the case that v 2 = v.) This
is one reason why definite integrals make better quiz questions than indefinite
integrals.
18.
Z
π
2
0
π
2
Z 1
Z 1
sin x
1
1
1
1
√
√
dx =
du =
+√
du
2
2 − cos2 x
2−u
2+u
0
0 2−u
0 2 2
√
√ 2 + u i1
√
1
1
2+1
1 h
= √ log(1 + 2).
= √ log √
= √ log √
2 2
2−u 0
2 2
2−1
2
sin x
dx =
1 + sin2 x
Z
19.
Z
9π
10
π
7
√
cosecx − sin x dx =
Z
=
Z
9π
10
π
7
9π
10
π
7
r
1
− sin x dx =
sin x
Z
9π
10
π
7
h √
i 9π
cos x
10
√
dx = 2 sin x π
sin x
7
s
1 − sin2 x
dx
sin x
!
r
r
π
9π
sin
=2
.
− sin
10
7
4
20.
Z
21.
Z 5
−5
|π − x| dx =
Z
cotx dx =
Z
cos x
dx = log | sin x| + C.
sin x
5
π − x dx +
Z
π
−5
π
h1
i5
h
1 iπ
+ x2 − πx
x − π dx = πx − x2
2
2
−5
π
25
25
1
1
= (π 2 − π 2 ) − (−5π − ) + ( − 5π) − ( π 2 − π 2 ) = π 2 + 25.
2
2
2
2
Alternatively, from a graph,
Z
5
−5
|pi − x| dx =
1
1
(π + 5)2 + (5 − π)2 = π 2 + 25.
2
2
22.
Z
e
dx =
Z
5 log x
dx =
Z
x5 dx =
x6
+ C.
6
23.
Z
ex +x
e
ex
e
x
· e dx =
Z
x
eu du = eu + C = ee + C.
24.
Z
sec x tan x dx = sec x + C.
(This is/used to be bookwork.)
25.
Z
log x
dx =
x
Z
1
du = log |u| + C = log | log x| + C.
u
26.
Z
x
1
√
+ C.
dx = sin−1
2
4 − x2
(This certainly is bookwork!
27.
Z
1
dx =
cos x + 1
Z
1
2
dt =
·
2
1 + t2
1−t
+1
1 + t2
Z
1 dt = t + C = tan−1
x
2
+ C.
5
28.
Z
cosec2 θ dθ = − cot θ + C.
This used to be bookwork once. Differentiate the right side to check!
29.
Z
30.
Z
Z
2
1
(3x2 )(6x)e3x dx =
18
2
1 2 3x2
3x e
− e3x + C =
=
18
2
x3 e3x dx =
tan
−1
x dx =
Z
Z
2
u sec u du = u tan u −
Z
1 u
1
ue du =
(ueu − eu ) + C
18
18
1 3x2
e (3x2 − 1) + C.
18
tan u du = u tan u + log | cos u| + C
= u tan u − log | sec u| + C = u tan u −
= x tan−1 x −
1
log(1 + tan2 u) + C
2
1
log(1 + x2 ) + C.
2
Alternatively, do an integration by parts with u = tan−1 x, dv = dx.
6
Group Stage
1.
sin(A + B) = sin A cos B + cos A sin B,
so
Z
π
4
sin 5x cos 3x dx =
π
8
Z
π
4
π
8
i π4
i π4
1h
1h
cos 8x π −
cos 2x π
16
4
8
8
√
2−1
.
=
8
=−
2.
Z
16π
2
2
sin x cos x dx =
0
= 2π.
Z
cos 8x cos 2x i π4
−
−
π
8
2
8
1
1
1
1
1
= − (1 − (−1)) −
0− √
=− + √
16
4
8 4 2
2
h1
1
(sin 8x + sin 2x) dx =
2
2
16π
1
sin2 2x dx =
4
0
Z
16π
0
h1
i16π
1
1
(1 − cos 4x) dx = x −
sin 4x
8
8
32
0
3.
Z
5
−5
|5 − x| dx =
5
i5
h 1
1
= 102 = 50.
5 − x dx = − (5 − x)2
2
2
−5
−5
Z
Look at the graph!
4.
Z
3
2
1
2
2 − 2x
√
dx =
2x − x2
Z
3
2
1
2
p
2(1 − x)
1 − (1 −
x)2
dx =
Z
− 21
1
2
√
2u
du = 0,
1 − u2
because the integrand is odd and the interval of integration is balanced about the
origin.
5.
Z
6.
Z
=
1
5
0
2
2
√
dx =
5
4 − 25x2
cos x
dx =
4 − sin2 x
1
log 3.
4
Z
1
0
Z
1
5
0
q
1
2 2
5
1
du =
4 − u2
Z
dx =
− x2
1
0
1
4
2 h −1 5x i 51
2 π
π
= · =
sin
.
5
2 0
5 6
15
1
1
+
2+u 2−u
du =
2 + u i1
1h
log 4
2−u 0
7
7.
Z
8.
Z 1
1
sin2 x cos2 x
e
e
dx =
0
Z
1
e dx = e.
0
1
i1
h1
1
1
1
−1 x − 1
−1
= − tan
dx =
tan
−
2
2
2
2 0
2
2
0 (x − 1) + 2
1
1
= tan−1
.
2
2
1
dx =
2
x − 2x + 5
0
Z
9.
Z
10.
Z π4
2
2
x3 ex dx =
1
Z
2
1
sin x sin 2x dx =
0
1
2
= √ = √ .
6 2
3 2
2
1
(2x)x2 ex dx =
2
Z
π
4
0
Z
4
1
h1
i4
1 u
3
ue du = (ueu − eu ) = e4 .
2
2
2
1
h sin x sin 3x i π4
1
1
1
(cos x − cos 3x) dx =
−
= √ − √
2
2
6
0
2 2 6 2
11.
Z
1
2
1
3
1
ex
dx =
x3
Z
3
i3
h
ueu du = ueu − eu = 2e3 − e2 .
2
2
12.
Z
13.
Z
e
5
cos(sin x) cos x dx =
2
Z
sin 5
cos u du = sin u
sin 2
(x2 − 3x + 2)−1 dx
Z
=e
14.
Z
1
x
x
e tan e dx =
0
h
isin 5
sin 2
= sin(sin 5) − sin(sin 2).
x − 1
1
1
x − 1
−
dx
log +C
x−1 x−2
x−2
=e
= eC .
x−2
Z
1
e
sec e ie
tan u du = log | sec u| = log .
sec 1
1
h
8
That’s what a blind calculation gives, but it is not correct, would you believe!
Sketch the graph of the integrand, and see what goes wrong! It makes you wonder
about some of the other integrals!
15.
1
Z
i1
h
xe−x dx = − xe−x − e−x = 1 − 2e−1 .
0
0
The indefinite integral can be guessed, or you can do an integration by parts.
16.
Z
x
20 x
(e + 1) e dx =
Z
e
(u + 1)20 du =
1
h1
i2
1
(u + 1)21 =
(e + 1)21 − 221 .
21
21
1
17.
Z
π
2
2
sin x cos x dx =
π
3
Z
1
2
u2 du =
0
h1
3
u3
i 21
0
=
1
.
24
18.
Z
19.
Z π4
1
0
h
i1
π
1
−1
dx
=
tan
x
= tan−1 1 = .
1 + x2
4
0
i π4
h
i π4
π
tan x + cot x dx = log | sec x| + log | sin x| π = log | tan x| π = − log tan
π
12
12
12
12
π
.
= log cot
12
h
20.
Z
2
1
1
dx =
x
e + e−x
Z
2
1
ex
dx =
e2x + 1
Z
e2
e
2
h
ie
1
−1
du
=
tan
u
u2 + 1
e
= tan−1 e2 − tan−1 e = tan−1
e2 − e
.
1 + e3
21
Z
b
a
1
sec2 log x dx =
x
Z
log b
log a
h
ilog b
= tan log b − tan log a.
sec2 u du = tan u
log a
9
22.
Z
1
sin
−1
x + cos
−1
x dx =
Z
1
0
0
π
π
dx = .
2
2
Draw the graph!
23.
Z
π
4
π
3
i π4
√
√
sec x dx = log | sec x + tan x| π = log( 2 + 1) − log(2 + 3) = log
h
3
!
√
2+1
√
3+2
!
√
3+2
√
,
2+1
= − log
which is negative, because of the unusual limits on the integral.
24.
Z
4
1
x3 − x
x
3
2
dx =
Z
4
3
1
x 2 − x− 2 dx =
1
h2
5
5
1
x 2 − 2x 2
i4
1
=
52
2
(32 − 1) − 2(2 − 1) =
.
5
5
25.
Z
π2
4
π2
9
√
Z π2
h
i π2
sin x
1
√
2 sin u du = − 2 cos u π = −2 0 −
dx =
= 1.
π
x
2
3
3
26.
Z
π
2
0
1
dx =
1 + cos x
Z
π
2
0
1
dx = 1.
1 + sin x
(Done earlier!)
27.
Z e
x + 49
dx =
x − 49
e
e − 49 h
ie
98
dx = x + 98 log |x − 49| = e − 1 + 98 log x
−
49
1
−
49
1
1
1
49 − e
.
= e − 1 + 98 log
48
Draw the graph!
Z
1+
10
28.
Z
7
log x
2
7
Z
dx =
xlog 2 dx =
5
5
h xlog 2+1 i7
1
=
7log 2+1 − 5log 2+1 .
log 2 + 1 5
log 2 + 1
29.
Z
π
2
0
1
dx =
sec x + tan x sin x
Z
0
π
2
cos x
dx =
1 + sin2 x
Z
1
0
h
i1
π
1
−1
du
=
tan
u
= .
2
1+u
4
0
30.
Z
e
1
1 − log x
dx =
x2
Z
0
1
1−t t
e dt =
e2t
Z
1
i1
h
(1 − t)e−t dt = te−t = e−1 .
0
0
31.
Z
π
3
4
sec x tan x dx =
0
Z
2
u3 du =
1
1
.
24
32.
1
Z
x2 (x3 + 1)4 dx =
0
Z
1
2
h u5 i2
31
1 4
=
u du =
.
3
15 1
15
33.
Z
2
π
sin2 x dx =
0
Z
2
π
0
34.
Z
π
e
h1
i π2
1 1
1
1
1
− cos 2x dx = x − sin 2x = − sin
2 2
2
4
π 4
0
cos 3x
dx =
3
4 cos x − 3 cos x
Z
π
1 dx = π − e.
e
(You have to know that cos 3x = 4 cos3 x − 3 cos x. Try to prove it.)
35.
2
π
Z
π
4
tan−1 θ + cot−1 θ dθ =
π
8
2
π
Z
π
4
1 dθ =
π
8
2 π
1
· = .
π 8
4
Unusual. tan−1 and cot−1 are usually applied to x, not θ.
4
.
π
11
36.
Z
π
4
etan x
dx =
cos2 x
0
Z
1
eu du = e − 1.
0
37.
Z √ π2
2
√ π x cos x dx =
6
Z
π
2
π
6
h1
i π2
1
1
cos u du =
sin u π =
2
2
2
6
1
1−
2
=
1
.
4
38.
Z
sec2 ( π
4 )−1
x
e dx =
Z
1
ex dx = 0.
1
1
39.
Z
log 3
0
ex
dx =
1 + ex
Z
4
2
1
du = log 4 − log 2 = log 2.
u
40.
Z
0
−1
2x + 1
dx =
2
x + 2x + 2
Z
1
0
h
i1
2u − 1
π
2
−1
du
=
log(u
+
1)
−
tan
u
= log 2 − .
2
u +1
4
0
41.
Z
12
4
Z √12
i√12
h
1
2
−1 u
2u du =
du = tan
(4 + u2 )u
4 + u2
2 2
2
2
√
π π
π
= tan−1 3 − tan−1 1 = − =
.
3
4
12
1
√ dx =
(4 + x) x
Z
√
12
42.
Z
3
1
2x2 − x + 1
dx
x−2
does not exist, because of the discontinuity at x = 2.
43.
Z
√
3
√
− 3
√
h
x i 3
π π 2π
−1
√
=− .
− −
dx = − sin−1
=
−
√
2 − 3
3
3
3
4 − x2
12
44.
Z
π
4
0
i π4
π 1
1
sin x dx = x − sin 2x = − .
2
4
8
4
0
h1
2
45.
Z
0
1
√
Z 1
Z 1
Z 1
u
2u2
2
x
dx =
2u
du
=
du
=
2− 2
du
2
2+1
1+x
1
+
u
u
u
+1
0
0
0
h
i1
π
= 2u − 2 tan−1 u = 2 − .
2
0
46.
Z
1
2
h
(x + 1)−2 dx = −
1 i2
1 1
1
=− + = .
x+1 1
3 2
6
47.
4
√
( x − 1)2 dx =
2
2
i2
4 3
4
2
u
−
u
+
u
√
√
√
2
3
2
2
2
2
√
√
4
1
4
8 8√
8 √
= (16 − 4) − (8 − 2 2) + (4 − 2) = 6 − (8 − 2 2 + 2 = − +
2 = ( 2 − 1).
2
3
3
3 3
3
Z
Z
2
(u − 1) 2u du =
Z
2u3 − 4u2 + 2u du =
h1
48.
i−1
2
3
x4 − x3 + x2 − 4x
4
3
2
2
2
2
3
1
=
(−1)4 − 24 −
(−1)3 − 23 +
(−1)2 − 22 − 4 (−1 − 2)
4
3
2
1
2
3
15
9
15 18
= × −15 − × −9 + × −3 − 4 × −3 = − + 6 − + 12 = 18 −
−
4
3
2
4
2
4
4
39
33
=
.
= 18 −
4
4
Z
−1
x3 − 2x2 + 3x − 4 dx =
h1
13
Quarter–Finals
1.
Z
√
1
3
sin x cos x
dx =
1
Z
p
= −√
tan
3
x/ sec4
x
dx =
Z
sec2 x
(tan x)
1
3
2
dx = −2(tan x)− 2 + C
2
+ C.
tan x
2.
Z
√
tan x
dx =
sin x cos x
√
tan x
dx =
tan x/ sec2 x
√
= 2 tan x + C.
Z
Z
secx
1
(tan x)
1
2
dx = 2(tan x) 2 + C
3.
Z
Z
sin5 x
sin x(1 − cos2 x)2
sin x(1 − 2 cos2 x + cos4 x)
dx =
dx =
dx
cos x
cos x
cos x
Z
1
= tan x − 2 cos x sin x + cos3 x sin x dx = log | sec x| + cos2 x − cos4 x + C.
4
Z
4.
Z
cos
√
x dx =
Z
√
√
√
2u cos u du = 2u sin u + 2 cos u + C = 2 x sin x + 2 cos x + C.
5.
1
1
− 2
2
x +1 x +2
dx =
3
3
1
= log(x2 + 1) − tan−1 x − log(x2 + 2) + √ tan−1
2
2
2
x
√
2
Z
3x − 1
dx =
2
(x + 1)(x2 + 2)
Z
(3x − 1)
Z
3x − 1 3x − 1
− 2
dx
x2 + 1
x +2
+ C.
6.
Z
x
dx =
4
x +1
Z
1 1
1
1
du = tan−1 u + C = tan−1 x2 + C.
2
2u +1
2
2
14
7.
Z
log cos x
dx =
cot x
Z
log u
−
du =
u
Z
1
2
1
cos6 x(1−cos2 x)2 sin x dx = − cos7 x+ cos9 x− cos11 x+C.
7
9
11
1
= − (log | cos x|)2 + C.
2
8.
Z
sin5 x cos6 x dx =
Z
1
1
−v dv = − v 2 + C = − (log |u|)2 + C
2
2
9.
Z
√
1
dx =
2
x + 2x
Z
1
1
√
dx =
du
2
2
u −1
(x + 1) − 1
Z
sec θ tan θ dθ = sec θ dθ = log | sec θ + tan θ| + C
Z
p
1
√
sec2 θ − 1
p
p
= log |u + u2 − 1| + C = log |x + 1 + x2 + 2x| + C.
=
Z
10.
Z
(3 + 2x) log x dx =
Z
t
t
(3 + 2e )te dt =
Z
3tet + 2te2t dt
1
1
= 3tet − 3et + te2t − e2t + C = 3x log |x| − 3x + x2 log |x| − x2 + C
2
2
1
= (x2 + 3x) log |x| − x2 − 3x + C.
2
11.
Z
12
dx =
2
(x + 4)(x2 + 16)
Z
1
1
1
1
−1 x
−1 x
−
+C.
−
dx
=
tan
tan
x2 + 4 x2 + 16
2
2
4
4
12.
√
If for the moment we pretend x2 = x, we find
Z p
Z p
Z
1 3
1
1 1
3
x2 + x4 dx = x 1 + x2 dx =
u 2 du = u 2 + C = (1 + x2 ) 2 + C
2
3
3
=
1
3
(x2 + x4 ) 2 + C.
3
3x
15
Luckily, since this function is odd (apart from the C), it is correct.
13.
1 1
+ cos 2θ dθ
2 2
1 1
1
1
1
1 p
= + sin 2θ + C = θ + sin θ cos θ + C = sin−1 x + x 1 − x2 + C.
2 4
2
2
2
2
Z p
1−
x2
dx =
Z p
2
1 − sin θ cos θ dθ =
Z
2
cos θ dθ =
Z
Actually, a similar comment applies to this integral as to the previous one!
14.
1
Z
dx =
3
x2 ) 2
1
Z
2
2
3
2
· 2 sec θ dθ =
(4 +
(4 + 4 tan θ)
Z
1
x
1
1
+ C.
=
cos θ dθ = sin θ + C = √
4
4
4 4 + x2
Z
1
· 2 sec2 θ dθ
8 sec3 θ
15.
Z
=
√
Z
1
e2x − 1
1
sec θ tan θ
= sec−1 ex + C.
Alternatively, = tan−1
Z
ex
1
√
√
dx =
du
x
2x
e e −1
u u2 − 1
Z
sec θ tan θ dθ = 1 dθ = θ + C = sec−1 u + C
dx =
√
Z
e2x − 1 + C.
16.
Z 2
1
1 + 2 cos 2x + cos2 2x dx
4
Z
Z
1
3 1
1
1 1
=
+ cos 2x + cos 4x dx
1 + 2 cos 2x + + cos 4x dx =
4
2 2
8 2
8
Z
4
cos x dx =
3
x+
8
3
= x+
8
3
= x+
8
=
1 1
+ cos 2x
2 2
dx =
Z
1
1
3
1
1
sin 2x +
sin 4x + C = x + sin x cos x +
sin 2x cos 2x + C
4
32
8
2
16
1
1
sin x cos x + sin x cos x(2 cos2 x − 1) + C
2
8
3
1
sin x cos x + sin x cos3 x + C.
8
4
16
17.
Z
√
=
Z
1
dx =
9 + 16x − 4x2
1
q
2
5 2
2
Z
1
1
q
q
dx =
dx
9
25
2
2
2 4 + 4x − x
2 4 − (x − 2)
2
1
(x − 2) + C.
dx = sin−1
2
5
Z
− (x − 2)2
18.
Z
sin log x dx =
Z
eu sin u du = eu (sin u − cos u) + C = x(sin log x − cos log x) + C.
(See question 9 of the Qualifying Round.)
19.
Z
2x tan
−1
2
x dx = x tan
−1
x−
Z
x2
dx = x2 tan−1 x −
1 + x2
= x2 tan−1 x − x + tan−1 x + C = (1 + x2 ) tan−1 x − x + C.
Z
1−
1
dx
1 + x2
20.
Z
Z
Z
2
1 ueu
1
1
(v − 1)ev−1
(v − 1)ev
x3 ex
dx
=
du
=
dv
=
dv
(x2 + 1)2
2 (u + 1)2
2
v2
2e
v2
Z
Z
1
1
1
1 ev
1
ev
v
v
v
v
=
+ e dv =
+C
(v − 1)e · − − − · ve dv =
−e +
2e
v
v
2e
v
2e v
Z
2
=
2
1 eu+1
1 eu
1 ex
ex
+C =
+C =
+
C
=
+ C.
2e u + 1
2u+1
2 x2 + 1
2(x2 + 1)
21.
Z
Z
Z
1 13 1 12
11
11
u − u +C
(x − 1)(x + 1) dx = u (u − 2) du = u12 − 2u11 du =
13
6
1
1
1
(x + 1)13 − (x + 1)12 + C =
(x + 1)12 (6(x + 1) − 13) + C
13
6
78
1
(x + 1)12 (6x − 7) + C..
=
78
=
17
22.
Z
1
1+x
1
4
dx =
Z
1
4u3 du =
1+u
Z
4
(v − 1)3 dv =
v
Z
1
4(v 2 − 3v + 3 − ) dv
v
1
3
4
= 4( v 3 − v 2 + 3v − log |v|) + C = (u + 1)3 − 6(u + 1)2 + 12(u + 1) − 4 log |u + 1| + C
3
2
3
1
1
1
4
4 3
= u3 − 2u2 + 4u − 4 log |u + 1| + C = x 4 − 2x 2 + 4x 4 − 4 log |1 + x 4 | + C.
3
3
23.
Z
sin5 (x−1 ) cos(x−1 )
dx =
x2
Z
1
1
− sin5 u cos u du = − sin6 u+C = − sin6 (x−1 )+C
6
6
24.
2
Z
√
2 cos x +
Z
sin
2 dx =
π
4
Z
π
π
dx = tan x +
+ C.
sec2 x +
4
4
25.
−1
x
2
= x sin−1 x
dx =
2
Z
u2 cos u du = u2 sin u + 2u cos u − 2 sin u + C
p
+ 2 1 − x2 sin−1 x − 2x + C.
18
Semi–Finals
1.
Z
sec2 x
dx =
sec x + tan x
The notoriously ugly
Z
Z
Z
2
sec x (sec x − tan x) dx =
Z
sec3 x dx −
1
tan2 x + C.
2
sec3 x dx can be calculated by integration by parts.
sec3 θ dθ =
1
1
sec θ tan θ + log | sec θ + tan θ| + C.
2
2
So our integral
1
1
1
sec θ tan θ + log | sec θ + tan θ| − tan2 θ + C
2
2
2
1
1
= tan θ(secθ − tan θ) + log | sec θ + tan θ| + C.
2
2
=
2.
Z
xx
e (log x+1)x
2x
dx =
Z
x
x
x
ueu du = ueu −eu +C = xx ex −ex +C = ex (xx −1)+C.
3.
√
Z
Z
2u2 − 9u + 9
4u3 − 18u2 + 18u
2x − 9 x + 9
du
√ 1 dx =
1 2u du =
1
(x − 3 x) 3
(u2 − 3u) 3
(u2 − 3u) 3
Z
Z
Z
2
2
(u2 − 3u)(4u − 6)
2
3
du = (u − 3u) 2(2u − 3) du = 2v 3 dv
=
1
2
(u − 3u) 3
√ 5
5
6
6
6 5
= v 3 + C = (u2 − 3u) 3 + C = (x − 3 x) 3 + C.
5
5
5
Z
4.
Z
Z
Z
1
1
x5
1 3
1 (u − 1)2
√
√
u 2 − 2u 2 + u− 2 du
du =
dx =
2
u
2
1 + x2
2 3
1
1 1
1 5
= u2 − u2 + u2 + C =
u 2 (3u2 − 10u + 15) + C
5
3
15
p
p
1
1
(3(x2 + 1)2 − 10(x2 + 1) + 15) 1 + x2 + C =
(3x4 − 4x2 + 8) 1 + x2 + C.
=
15
15
19
5.
log(1 + log x)
dx =
x
Z
Z
log u du = u log u − u + C
= (1 + log x) log(1 + log x) − log x + C.
6.
Z
1
x2
1
3
1
4
dx =
Z
u6
12u11 du =
u4 + u3
Z
12u14
du =
u+1
Z
12(u14 − 1) + 12
du
u+1
x +x
Z
= 12 u13 − u12 + u11 − u10 + u9 − u8 + u7 − u6 + u5 − u4
1 du
u+1
12
12
6
4
3
12
12
6
= u14 − u13 + u12 − u11 + u10 − u9 + u8 − u7 + 2u6 − u5
7
13
11
5
3
2
7
5
+ u3 − u2 + u − 1 +
+ 3u4 − 4u3 + 6u2 − 12u + 12 log |u + 1| + C
1
Now set u = x 12 .
7.
Z
x3 + 3x2 + 3x
dx =
x4 + 4x3 + 6x2 + 4x + 1
= log |x + 1| +
1
+ C.
3(x + 1)3
Z
(x + 1)3 − 1
dx =
(x + 1)4
Z
1
1
−
dx
x + 1 (x + 1)4
8.
Z
sec x cosecx dx =
Z
1
dx =
cos x sin x
= − log |cosec 2x + cot 2x| + C.
9.
Z
1
dx =
x + x4
2
dx =
sin 2x
Z
2cosec2x dx
A
B
Cx + D
+
+ 2
dx
x
x+1 x −x+1
C
C
2
1
= A log |x| + B log |x + 1| + log(x2 − x + 1) + D +
sin−1 √
x−
+ K,
2
2
2
3
Z
1
dx =
x(x + 1)(x2 − x + 1)
Z
Z
20
where A, B, C and D are given by
A(x + 1)(x2 − x + 1) + Bx(x2 − x + 1) + (Cx + D)x(x + 1) = 1
(for all x).
We have
A + B + C = 0, −B + C + D = 0, B + D = 0, A = 1,
so
1
2
1
A = 1, B = − , C = − , D = ,
3
3
3
and the integral is
1
1
1
log |x + 1| − log(x2 − x + 1) + K = log |x| − log |x3 + 1| + K
3
3
3
3 4
1
1
x x = log 3
+ K.
+ K = log 3
x +1
3
x + x4
= log |x| −
10.
Z
cosec x dx = − log |cosec x + cot x| + C.
(Pure bookwork.)
11.
Z
cos sin sin x cos sin x cos x dx = sin sin sin x + C.
(Obvious!)
12.
Z
Z
x
u−1
x
p
√
dx =
du
dx =
1 − 2x − x2
2 − u2
2 − (x + 1)2
p
p
u
x+1
−1
−1
2
2
√
√
= − 2 − u − sin
+ C = − 1 − 2x − x − sin
+ C.
2
2
Z
13.
Z
√
Z
1 4v
1 1 4v
1 4v
1 3
u log u du =
ve dv = 4
ve − e
x log(5x) dx =
+C
54
54
5
4
16
1
1
1
1 1 4
u log u − u4 + C = x4 log 5x − x4 + C.
= 4
5
4
16
4
16
3
Z
21
14.
Z log x
x
2
dx =
Z 2
Z
u
u
e
du
=
u2 e−u du
eu
= −u2 e−u − 2ue−u − 2e−u + C = −
(log x)2
log x 2
−2
− + C.
x
x
x
15. If n is an integer, n ≥ 2,
Z
Z
1
xn−2
1
dx
=
dx
=
dx
xn + x
x(xn−1 + 1)
xn−1 (xn−1 + 1)
Z
1
1
n−2
= x
− n−1
dx
xn−1
x
+1
Z
1
xn−2
1
=
− n−1
dx = log |x| −
log |xn−1 + 1| + C.
x x
+1
n−1
Z
22
Finals
1.
Z
=
2
4
sin x cos x dx =
1
8
Z 1
sin 2x
2
2 1 1
+ cos 2x
2 2
1
dx =
8
Z
sin2 2x + sin2 2x cos 2x dx
1
1
1
1 1
− cos 4x + sin2 2x cos 2x dx =
x−
sin 4x +
sin3 2x + C.
2 2
16
64
48
Z
This solution was indicated to me by one of the competitors.
2.
4x2 − 15x + 29
dx =
(x − 5)(x2 − 4x + 13)
A
Bx + C
+
dx
x − 5 (x − 2)2 + 32
B
x−2
2B + C
2
−1
= A log |x − 5| + log(x − 4x + 13) +
tan
+ K,
2
3
3
Z
Z
where A, B and C are given by
A(x2 − 4x + 13) + (Bx + C)(x − 5) = 4x2 − 15x + 29
(for all x).
We have
A + B = 4, −4A − 5B + C = −15, 13A − 5C = 29,
so
A = 3, B = 1, C = 2
and the integral is
1
4
= 3 log |x − 5| + log(x2 − 4x + 13) + tan−1
2
3
3.
Z
Z √
x−2
3
+ K.
Z √
1 + sin x
1+u
cos x dx =
du
sin x
u
Z
Z
Z
Z
v
1
2v 2
2
1
=
2v dv =
dv = 2 + 2
dv = 2 +
−
dv
v2 − 1
v2 − 1
v −1
v−1 v+1
√1 + u − 1 v − 1
√
= 2v + log + C = 2 1 + u + log √
+C
v+1
1+u+1
√1 + sin x − 1 √
= 2 1 + sin x + log √
+ C.
1 + sin x + 1
√
1 + sin x cot x dx =
23
4.
Z
Z
sin 2x
1 1
1
sin 2x
dx = −
du = log |2 − u| + C
dx =
2
2
−
cos
2x
2
2
−
u
2
1 + 2 sin x
1
1
= log |2 − cos 2x| + C = log(2 − cos 2x) + C.
2
2
Z
5.
Z
Z
Z
x4 − 1 + 1
1
x4
2
dx
=
−
dx
=
−
x
+
1
+
dx
x2 − 1
x2 − 1
x2 − 1
Z
1
1
1
1
2
+
dx = −x − 1 +
dx
1 − x2
2 1−x 1+x
1 + x
1
1
= − x3 − x + log +C
3
2
1−x
x4
dx =
1 − x2
Z
= −x2 − 1 +
Z
−
6.
Z √
Z
Z p
cos2 θ
1−x
1 − sin2 θ
√ dx =
2 sin θ cos θ dθ =
2 sin θ dθ
1 − sin θ
1 − sin θ
1− x
Z
Z
1
= (1 + sin θ)2 sin θ dθ = 2 sin θ + 1 − cos 2θ dθ = −2 cos θ + θ − sin 2θ + C
2
√
√
√
√
= θ − 2 cos θ − sin θ cos θ + C = sin−1 x − 2 1 − x − x 1 − x + C.
7.
Z
cot x+2 log cosec x
e
dx =
Z
cot x
e
2
cosec x dx =
Z
−eu du = −eu + C = −ecot x + C.
8.
Z
−1
xesin x
1
√
dx = eθ sin θ dθ = (sin θ − cos θ)eθ + C
2
1 − x2
p
−1
1
x − 1 − x2 esin x + C
=
2
Z
9.
Z 12
5
(2 sin x + 3 cos x) − 13
(2 cos x − 3 sin x)
3 sin x + 2 cos x
13
dx =
dx
2 sin x + 3 cos x
2 sin x + 3 cos x
Z
12
5 2 cos x − 3 sin x
12
5
=
−
dx =
x−
log |2 sin x + 3 cos x| + C.
13 13 2 sin x + 3 cos x
13
13
Z
24
10.
Z
Z
1
1
1
dx =
dx =
dx
sin x + sin 2x
sin x + 2 sin x cos x
sin x(1 + 2 cos x)
Z
Z
1
1
=
sin x dx = −
du
2
2
(1 − cos x)(1 + 2 cos x)
(1 − u )(1 + 2u)
Z
1 1
4 1
1 1
−
−
du
=
2 1 + u 6 1 − u 3 1 + 2u
Z
1
1
2
log |1 + u| + log |1 − u| − log |1 + 2u| + C
2
6
3
1
2
1
= log |1 + cos x| + log |1 − cos x| − log |1 + 2 cos x| + C
2
6
3
2
3
1
(1 + cos x) (1 − cos x)
sin x(1 + cos x)2
1
= log
+ C = log
+C
6
(1 + 2 cos x)4
6
(1 + 2 cos x)4
sin x(1 + cos x) 1
= log +C
3
(1 + 2 cos x)2
=
25
Tie–breaker questions
1.
Let I =
π
2
Z
√
0
Then
q
0
2I =
Z
sin x
dx.
√
sin x + cos x
q
sin π2 − x
q
− x + cos
π
2
Z
I=
√
π
2
1 dx =
0
sin
π
2
π
2
−x
π
π
, so I = .
2
4
dx =
Z
0
π
2
√
√
cos x
√
dx,
cos x + sin x
2.
1
Z 12
h
i 12 Z 12
u
sin−1 log x
−1
−1
√
sin u du = u sin u −
dx =
du
x
0
1 − u2
0
0
1
√
hp
i 21
3
π
π
+
+
− 1.
1 − u2 =
=
12
12
2
0
Z
e2
Z
1
3.
x7 − 1
dx =
log x
0
=
Z
7
Z
1
0
7
Z
y
x dy dx =
0
Z
7
0
Z
0
1
y
x dx dy =
Z
7
y+1
0
h
i7
1
dy = log(y + 1) = log 8.
y+1
0
0
h xy+1 i1
0
dy
Clearly not suitable for first–year students.
4.
Let I =
π
2
Z
1
dx.
1 + tanπ x
0
Z
Then I =
and 2I =
Z
π
2
0
1
π
2
1 + tanπ
π
2
1 dx =
0
π
,
2
−x
I=
dx =
Z
0
π
2
1
dx =
1 + cotπ x
Z
0
π
2
tanπ x
dx,
tanπ x + 1
π
.
4
5.
Z
0
π
6
3
sec 2θ dθ =
Z
0
π
3
h1
i π3
1
1
sec3 u du =
sec θ tan θ + log | sec θ + tan θ|
2
4
4
0
√
√
3 1
+ log(2 + 3).
=
2
4

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