MATH6501 - Autumn 2016
Solutions to Problem Sheet 5
1. (a) Let
u = x2 + 2
⇒
du = 2x dx.
Then
Z
Z
p
1
1
3
2
x
x + 2 dx =
(u − 2)u 2 du
2
Z
1
1 3
=
u 2 − 2u 2 du
2
1 2 5 4 3
=
u2 − u2 + C
2 5
3
1 5 2 3
= u2 − u2 + C
5
3
5 2 2
3
1 2
=
x +2 2 −
x + 2 2 + C.
5
3
(b) Let
u = x3
⇒
du = 3x2 dx,
hence
Z
2
Z
3
x cos(x ) dx =
1
cos u du
3
1
sin u + C
3
1
= sin (x3 ) + C.
3
=
(c) Let
u=
1
x
⇒
du = −
1
dx,
x2
MATH6501 - Autumn 2016
Z
1
sin
x2
Z
1
dx = − sin u du
x
= cos u + C
1
+ C.
= cos
x
(d) Let
u = 3x − 7
⇒ du = 3 dx,
Z
Z
1
cos (3x − 7) dx =
cos u du
3
1
= sin u + C
3
1
= sin (3x − 7) + C.
3
(e) Let
u=
√
x
⇒
1
1
du = √ dx =
dx
2u
2 x
√
Z
e x
√ dx =
x
Z
eu
2
u du
u
Z
2eu du
=
= 2eu + C
√
=e
x
+ C.
(f) As stated in the question, we will use the following
substitution. . .
u = ex
⇒
du = ex dx = u dx.
MATH6501 - Autumn 2016
Also, note that
x)
e(x+e
x)
= ex e(e
= ueu ,
therefore
Z
x
e(x+e ) dx =
Z
ue
Z
=
u1
u
du
eu du
= eu + C
x
= e(e ) + C.
2. (a) Start by completing the square. . .
Z
Z
2
2
dx
=
dx
x2 − 6x + 10
(x − 3)2 + 1
Z
2
=
du,
2
u +1
where
⇒
u=x+3
du = dx.
Then the denominator of (u2 + a2 ) with a = 1 suggests we should substitute:
u = tan θ
⇒
du = sec2 θdθ.
MATH6501 - Autumn 2016
This yields
Z
Z
1
2
du = 2
u2 + 1
Z
=2
Z
=2
1
tan2 θ
+1
sec2 θdθ
1
sec2 θdθ
2
sec θ
dθ
= 2θ + C
= 2 tan−1 u + C
= 2 tan−1 (x + 3) + C.
Note: This integral can be tackled with just one
substitution if you let
tan θ = x + 3
⇒
sec2 θdθ = dx.
(b) Start by completing the square:
Z
Z
1
1
√
p
dx =
dx
2
−x + 4x − 3
1 − (x − 2)2
Z
1
√
du,
=
1 − u2
with
u = x − 2 ⇒ du = dx.
√
Then the factor of 1 − u2 suggests that we should
make a second substitution u = a sin θ with a = 1,
i.e.
u = sin θ ⇒ du = cos θ dθ,
MATH6501 - Autumn 2016
and so we can rewrite the integral as
Z
Z
1
1
√
p
cos θ dθ
du =
2
1−u
1 − sin2 θ
Z
1
=
cos
θ dθ
cos
θ
Z
= dθ
=θ+C
= sin−1 u + C
= sin−1 (x − 2) + C.
(c) Observe that
Z
Z
2
2
√
p
dx
dx =
4x2 − 9
(2x)2 − 32
Z
2
q
=
dx
3 2
2
2 x − 2
Z
1
q
=
2 dx,
x2 − 23
which contains a factor of
So substitute
x=
3
cosh θ
2
⇒
√
x2 − a2 where a = 3/2.
dx =
3
sinh θ dθ
2
MATH6501 - Autumn 2016
into the integral to obtain
Z
Z
2
3
2
√
p
sinh θ dθ
dx =
2
2
2
4x − 9
(2x) − 3 2
Z
1
3
q
=
sinh θ dθ
2
2
2
3
3
cosh
θ
−
2
2
Z
3
1
sinh θ dθ
p
=
2
3
2
cosh
θ
−
1
Z 2
sinh θ
p
dθ,
=
cosh2 θ − 1
and since cosh2 θ − sinh2 θ ≡ 1,
Z
Z
2
sinh
θ
√
dθ
dx =
2
sinh
θ
4x − 9
Z
=
dθ
=θ+C
= cosh−1
2x
3
+ C.
3. (a) If one has f (x) = cos x, then we can make the
MATH6501 - Autumn 2016
following observation. . .
Z
Z
− sin x
tan x dx = −
dx
cos x
Z 0
f (x)
=−
dx
f (x)
= ln |f (x)| + C
= ln |cos x| + C
(b) Using
sin2 x + cos2 x ≡ 1
we have
Z
⇒
Z
5
sin x dx =
Z
=
sin2 x ≡ 1 − cos2 x,
sin4 x sin x dx
2
1 − cos2 x sin xdx.
Looking at the rewritten integral, we would like to
have a substitution that gives
du = ± sin xdx,
which suggests we should choose
u = cos x
⇒
du = − sin xdx.
MATH6501 - Autumn 2016
Then
Z
Z
2
5
sin x dx = −
1 − u2 du
Z
=
−u2 + 2u2 − 1 du
1
2
= − u5 + u3 − u + C
5
3
2
1
5
= − cos x + cos3 x − cos x + C.
5
3
(c) Note that
Z
1
dx =
x ln x
Z
1
x
dx
ln x
Z 0
f (x)
dx,
=
f (x)
for f (x) = ln x. Hence
Z
1
dx = ln |f (x)| + C
x ln x
= ln |ln x| + C
4. To find the expected money made, evaluate
Z 3
Z 3
dr
2
dx =
2−
dx
(x + 1)2
0 dx
0
3
2
= 2x +
x+1 0
2
2
= 2·3+
− (2 · 0 +
)
3+1
0+1
= 6.5 − 2
= 4.5,
MATH6501 - Autumn 2016
i.e. £4500 for 3000 whisks.