MA1S11 Tutorial Sheet 10, solutions1
15-18 December 2015
Questions
1. (4) Determine the following indefinite integrals
√
Z
(1 + x)20
√
dx
a)
x
Z
b)
tan(2x) sec2 (2x)dx
Z
√
c)
x2 3 + xdx
Solution: a) is a u-substitution; let u = 1 +
so
du =
√
x, now
1
du
= √
dx
2 x
(1)
1
du
dx = √ dx
dx
2 x
(2)
and so the integral become
√
√
Z
Z
2(1 + x)21
(1 + x)20
2u21
20
√
+C =
+C
dx = 2 u du =
21
21
x
Again with the u-substitution for b), let u = tan(2x) so du = 2 sec(2x)dx and
Z
Z
1
u2
tan2 (2x)
2
tan(2x) sec (2x)dx =
udu =
+C =
+C
2
4
4
(3)
(4)
Note that one may also use the substitution u = sec(2x), then du = 2 tan(2x) sec(2x)dx
and
Z
Z
1
u2
sec2 (2x)
2
tan(2x) sec (2x)dx =
udu =
+C =
+C
(5)
2
4
4
There is no contradiction here as
tan2 (2x) = sec2 (2x) − 1,
(6)
so the difference is a constant which is part of the ambiguity in the antiderivative
encoded in the constant C.
1
Stefan Sint, sint@maths.tcd.ie, see also http://www.maths.tcd.ie/~sint/MA1S11.html
1
For b), use u = 3 + x so du = dx but we have to substitute x = u − 3 back in to the
equation, the equation has x2 in it and
x2 = (u − 3)2 = u2 − 6u + 9
(7)
giving
Z
x
3
√
Z
3 + xdx =
Z
=
√
(u2 − 6u + 9u) udu
(u5/2 − 6u3/2 + 9u1/2 )du
2 7/2
2
2
u − 6 u5/2 + 9 u3/2 + C
7
3
5
√
12
2
3
2
=
(3 + x) − (3 + x) + 6(3 + x) + C
3+x
7
5
=
(8)
where we could multiply out the cubic polynomial in x (if we felt like it).
2. (4) Calculate the definite integrals
Z
π/2
a)
cos(x)dx
0
Z
2
b)
Z1 5
c)
2
Z
−2
y − y dy
and
1
y 2 − y −2 dy
2
(1 + w)(2w + w2 )5 dw
−1
Solution: Well a) should be straight forward enough, we know the antiderivative of
cos x is sin x and hence
Z π/2
π π/2
cos(x)dx = sin(x)|0 = sin
− sin(0) = 1 − 0 = 1
(9)
2
0
and for b), we know the antiderivative of y 2 is y 3 /3 and of y −2 , −y −1 , hence
2
Z
1
and
Z
2
3
2
y
1
8 1 1
11
+
y − y dy =
= + − −1=
3
y 1 3 2 3
6
(10)
3
1
y
1
1
8 1
11
y − y dy =
+
= +1− − =−
3
y 2 3
3 2
6
(11)
2
1
2
−2
−2
as expected by the rule that an exchange of the limits of integration is equivalent to
a change of sign.
2
In c) let u = 2w + w2 so that du = (2 + 2w)dw = 2(1 + w)dw. When w = −1 by
substituting back in we have u = −1 and when w = 5, u = 35, hence
35
Z 5
Z
1 35 5
u6 356 1
2 5
(1 + w)(2w + w ) dw =
u du =
−
(12)
=
2 −1
6 −1
6
6
−1
Extra Questions
The questions are extra; you don’t need to do them in the tutorial class.
3. Calculate using u-substitution
Z
π/2
cos(x) cos (sin(x))dx
−π
Solution: Let u = sin(x) so du = cos(x)dx. When x = −π we have u = 0 and when
x = π/2 we have u = 1 giving
π/2
Z
Z
cos(x) cos (sin(x)) dx =
−π
1
cos(u)du = sin(u)|10 = sin(1)
(13)
0
4. Calculate the definite integrals
Z
π/2
a)
sin(x) cos(x)dx
0
Z
π/4
b)
sin(x) cos(x)dx
0
Z
π/2
sin(x) cos(x)dx
c)
π/4
and verify the property of the definite integral
Z b
Z c
Z b
f (x)dx =
f (x)dx +
f (x)dx
a
a
c
with f (x) = sin(x) cos(x), a = π/4, b = 0, c = π/2.
Solution: We can either use u-substitution e.g. u = sin(x) so that du = cos(x)dx,
or we use a trigonometric identity to first simplify the integrand sin(x) cos(x) =
1
sin(2x). One then finds
2
Z
a
b
b
1
1
sin(x) cos(x)dx = − cos(2x) = − (cos(2a) − cos(2b))
4
4
a
3
(14)
This evaluates to a) 1/2, b) 1/4 and c) 1/4. To check the additive property of the
definite integral we evaluate first the LHS:
Z
b
Z
0
sin(x) cos(x)dx = −
sin(x) cos(x)dx =
a
Z
π/4
π/4
sin(x) cos(x)dx = −
0
1
4
(15)
where the last equality uses the result in b). For the RHS we have
Z
c
Z
sin(x) cos(x)dx+
a
b
Z
π/2
sin(x) cos(x)dx =
c
Z
0
sin(x) cos(x)dx+
π/4
sin(x) cos(x)dx
π/2
(16)
Exchanging the limits of integration in the second integral and using the results of
c) and a) we get
Z
π/2
Z
π/2
sin(x) cos(x)dx −
π/4
sin(x) cos(x)dx =
0
which is equal to the LHS, as expected.
4
1 1
1
− =−
4 2
4
(17)