MA1311 (Advanced Calculus) Tutorial/Exercise sheet 10
[due 17.50am, Tuesday, December 14, 2010]
Brief solutions
Z
1. Find
(x2 − 3x)e4x dx
Solution: Integration by parts (twice) does this.
Z
(x2 − 3x)e4x dx
Let u = x2 − 3x
dv = e4x dx
du = (2x − 3) dx v = 41 e4x
Z
Z
(x2 − 3x)e4x dx =
u dv
Z
= uv − v du
Z
1 4x
1 4x
2
= (x − 3x)
e
e (2x − 3) dx
−
4
4
Let U = 2x − 3 dV = 14 e4x dx
1 4x
dU = 2 dx
V = 16
e
Z
2
x − 3x 4x
e − U dV
=
4
Z
x2 − 3x 4x
=
e − (U V − V dU )
4
Z
x2 − 3x 4x
=
e − U V + V dU
4
Z
x2 − 3x 4x 2x − 3 4x
1 4x
=
e −
e +
e 2 dx
4
16
16
x2 − 3x 4x 2x − 3 4x
1
=
e −
e + e4x + C
4
16
32
2
8x − 28x + 7 4x
e +C
=
32
x2 − 3x
2. Find
dx
2x3 − 9x2 + 21
Solution: Substitution of u = 2x3 − 9x2 + 21, du = (6x2 − 18x) dx = 6(x2 − 3x) dx
Z
works here
x2 − 3x
dx =
2x3 − 9x2 + 21
du
x2 − 3x
u
6(x2 − 3x)
Z
11
=
du
6u
1
ln |u| + C
=
6
1
=
ln |2x3 − 9x2 + 21| + C
6
Z
Z
3. Find
Z
cos4 (3x) dx
Solution: Method for even powers of cos θ is to use cos2 θ = 21 (1 + cos 2θ)
Z
Z
4
cos (3x) dx =
(cos2 (3x))2 dx
2
Z 1
=
(1 + cos(6x) dx
2
Z
1
=
(1 + 2 cos(6x) + cos2 6x) dx
4
Z 1
1
=
1 + 2 cos(6x) + cos(12x) dx
4
2
Z 3 1
1
=
+ cos(6x) + cos(12x) dx
8 2
8
3
1
1
=
x+
sin(6x) +
sin(12x) + C
8
12
96
Z
π/4
sec2 x dx
4. Find
0
Solution:
Z
π/4
π/4
sec2 x dx = [tan x]0
= tan(π/4) = tan 0 = 1 − 0 = 1
0
Z
5. Find
tan2 x dx
Solution:
Z
2
tan x dx =
Z
sec2 x − 1 dx = tan x − x + C
2
Z
6. Find
cos2 x sin3 x dx
du
,
Solution: Odd power of sin x, so substitute u = cos x, du = − sin x dx, dx = − sin
x
Z
Z
−du
2
3
cos x sin x dx =
u2 sin3 x
sin x
Z
=
−u2 sin2 x du
Z
=
−u2 (1 − cos2 x) du
Z
=
−u2 (1 − u2 ) du
Z
=
u4 − u2 du
1 5 1 3
u − u +C
5
3
1
1
=
cos5 x − cos3 x + C
5
3
=
x2
dx
x+1
Solution: Not really a case for partial fractions, but covered by the general method for
partial fractions. This is not a proper fraction. So do long division
Z
7. Find
x − 1
x + 1 x2
x2 + x
− x
− x − 1
1
Thus we have
x2
remainder
1
= quotient +
=x−1+
x+1
x+1
x+1
Z Z
2
x
1
1
dx =
x−1+
dx = x2 − x + ln |x + 1| + C
x+1
x+1
2
3
Z
8. Find
π/6
e3x cos(2x) dx
0
Solution: Integration by parts works here but it is tricky. Need to do it twice.
Z
π/6
e3x cos(2x) dx
Let u = e3x
dv = cos(2x) dx
3x
0
du = 3e dx v = 12 sin(2x)
Z π/6
Z π/6
3x
e cos(2x) dx =
u dv
0
0
Z π/6
π/6
= [uv]0 −
v du
0
π/6 Z π/6
1
1 3x
e sin(2x)
−
sin(2x)(3e3x ) dx
2
2
0
0
Z
1
3 π/6 3x
1 π/2
e sin(π/3) − sin 0 −
e sin(2x) dx
2
2
2 0
Let U = e3x
dv = sin(2x) dx
3x
dU = 3e dx V = − 12 cos(2x)
√
Z
3 π/2 3 π/6
U dV
e −
4
2 0
!
√
Z π/6
3 π/2 3
π/6
[U V ]0 −
V dU
e −
4
2
0
!
√
π/6 Z π/6 1
1 3x
3 π/2 3
− e cos(2x)
e −
−
− cos(2x) (3e3x ) dx
4
2
2
2
0
0
√
Z
9 π/6 3x
3 π/2 3 3x
π/6
e + [e cos(2x)]0 −
e cos(2x) dx
4
4
4 0
√
Z
3 π/2 3 π/2 1
9 π/6 3x
e +
e
−1 −
e cos(2x) dx
4
4
2
4 0
=
=
=
=
=
=
=
R π/6
We appear to be back to where we started, which is looking for 0 e3x cos(2x) dx, but
we actually have an equation now that must be satisfied by this number.
R π/6
Let I = 0 e3x cos(2x) dx, and think of it as an unknown number. We have shown above
that
√
3 π/2 3 π/2 3 9
I=
e + e − − I
4
8
4 4
That says
√
9
3 π/2 3 π/2 3
1+
I=
e + e −
4
4
8
4
4
and so
4
I=
13
√
3 π/2 3 π/2 3
e + e −
4
8
4
√
!
=
3 π/2
3
3
e + eπ/2 −
13
26
13
Z
1
dx
+x+1
Solution: The denominator has complex roots (can’t be factored without using complex
numbers) and so we do this by completing the square
9. Find
x2
1 1
x +x+1=x +x+ − +1=
4 4
2
2
1
x−
2
2
+
3
4
√
It is possible to do a substitution x − 12 = 23 tan θ or to know that the answer is
Z
Z
1
1
dx =
dx
2
2
x +x+1
x − 12 + 43
2
x − 1/2
−1
√
= √ tan
+C
3
3/2
2
2x − 1
−1
√
= √ tan
+C
3
3
x2
dx
(x + 2)(x2 − 1)
Solution: Partial fractions. Proper fraction (denominator of degree 3, numerator of degree
2 < 3). Factorise denominator completely (x + 2)(x2 − 1) = (x + 2)(x − 1)(x + 1). Partial
fractions take the form
Z
10. Find
x2
A1
A2
A3
x2
=
=
+
+
(x + 2)(x2 − 1)
(x + 2)(x − 1)(x + 1)
x+2 x−1 x+1
Can work out A1 , A2 and A3 by multiplying across by (x + 2)(x − 1)(x + 1) and then
substituting x = −2, x = 1 and x = −1 in turn.
Result is
Z
x2
dx =
(x + 2)(x2 − 1)
Z 1/6
1/2
4/3
+
−
dx
x+2 x−1 x+1
4
1
1
=
ln |x + 2| + ln |x − 1| − ln |x + 1| + C
3
6
2
5
x2
dx
(x + 1)(x2 − 1)
Solution: Partial fraction again. Again a proper fraction. Factorise denominator completely (x + 1)(x2 − 1) = (x + 1)(x − 1)(x + 1) = (x + 1)2 (x − 1). Partial fractions take
the form
Z
11. Find
x2
A2
x2
A1
A3
=
=
+
+
(x + 1)(x2 − 1)
(x + 1)2 (x − 1)
x + 1 (x + 1)2 x − 1
Can work out A1 , A2 and A3 by multiplying across by (x+1)2 (x−1) and then substituting
x = −1, x = 1 and x = 0 in turn.
Result is
Z
x2
dx =
(x + 1)(x2 − 1)
Z 1/4
3/4
1/2
+
−
dx
x + 1 (x + 1)2 x − 1
3
1
1
=
ln |x + 1| +
+ ln |x − 1| + C
4
2(x + 1) 4
x2
dx
(x + 1)(x2 + 1)
Solution: Partial fraction again. Again a proper fraction. Denominator is already factorised
completely. Partial fractions take the form
Z
12. Find
A
Bx + C
x2
=
+ 2
2
(x + 1)(x + 1)
x+1
x +1
Can work out A by multiplying across by (x + 1)(x2 + 1) and then substituting x = −1.
Substituting x = 0 finds C and another x then gives B.
Result is
Z
x2
dx =
(x + 1)(x2 + 1)
Z 1/2
(1/2)x − 1/2
+
dx
x+1
x2 + 1
Z 1
x
1
=
ln |x + 1| +
−
dx
2
2(x2 + 1) 2(x2 + 1)
1
1
1
=
ln |x + 1| + ln(x2 + 1) − tan−1 x + const
2
4
2
(I used ‘+ const’ rather than the usual ‘+C’ just because I used C already for something
else.)
Richard M. Timoney
6