Unit #10 - Graphs of Antiderivatives; Substitution Integrals
Section 7.1
Some material from “Calculus, Single and MultiVariable” by Hughes-Hallett, Gleason,
McCallum et. al.
Copyright 2005 by John Wiley & Sons, Inc.
This material is used by permission of John Wiley & Sons, Inc.
TEST PREPARATION PROBLEMS
Note: In the solutions for #3-#40, we always show the substitution used. On a test, if you can
compute the anti-derivative in your head, you do not need to go through all the steps shown here. They
are included in these solutions as learning & comprehension aid.
Z
2
3.
tet dt
1
dw
Let w = t2 , so dw = 2t dt or dt =
2t
Z
Z
Z
2
1 w
1
1 2
1
dw =
e dw = ew + C = et + C
tet dt = tew
2t
2
2
2
2
2
d 1 t2
1
Check:
e + C = (2t)et = tet = original function in integral.
dt 2
2
4.
Z
"
e3x dx
dw
Let w = 3x, so dw = 3 dx or dx =
3
Z
Z
1 w
1 3x
3x
w dw
= e +C = e +C
e dx = e
3
3
3
d 1 3x
1 3x
Check:
e + C = e (3) = e3x = original function in integral.
dx 3
3
NOTE: we will not show the differentiation check for any later questions, as the
process is always the same, and you should be comfortable enough with the derivative
rules to do the checking independently. If you are uncertain about any problem,
contact your instructor.
Z
5.
e−x dx
"
Z
6.
Z
Let w = −x, so dw = −1 dx or dx = (−dw)
Z
1
e−x dx = ew (−dw) = ew + C = −e−x + C
3
25e−0.2t dt
dw
Let w = −0.2t, so dw = −0.2 dt or dt =
= −5dw
−0.2
Z
Z
25e−0.2t dt = 25ew (−5dw) = −125ew + C = −125e−0.2t + C
1
7.
Z
t cos(t2 ) dt
dw
Let w = t2 , so dw = 2t dt or dt =
2t
Z
Z
1
1
dw
2
= sin(w) + C = sin(t2 ) + C
t cos(t ) dt = t cos(w)
2t
2
2
8.
Z
sin(2x) dx
1
Let w = 2x, so dw = 2dx, or dx = dw
2
Z
Z
1
1
1
sin(w)dw = − cos(w) + C = − cos(2x) + C
sin(2x)dx =
2
2
2
9.
Z
sin(3 − t) dt
Z
10.
Z
Let w = 3 − t, so dw = −1 dt or dt = −dw
Z
sin(3 − t) dt = sin(w) (−1)dw = −(− cos(w)) + C = cos(3 − t) + C
2
xe−x dx
−1
Let w = −x2 , so dw = −2x dx, or dx =
dw
2x
Z
Z
Z
2
2
−1 w
1
1
−1
dw =
e dw = − ew + C = − e−x + C
xe−x dx = xew
2x
2
2
2
11.
Z
(r + 1)3 dr
Let w = (r + 1)3 , so dw = dr
Z
Z
w4
(r + 1)4
3
(r + 1) dr = w3 dw =
+C =
+C
4
4
12.
Z
y(y 2 + 5)8 dy
1
dw
Let w = y 2 + 5, so dw = 2y dy, or dy =
2y
Z
Z
Z
1
1
1 2
1 w9
y(y 2 + 5)8 dy = yw8
dw =
+C =
(y + 5)9 + C
w8 dw =
2y
2
2 9
18
13.
Z
t2 (t3 − 3)10 dt
1
Let w = t3 − 3, so dw = 3t2 dt, or dt = 2 dw
3t
Z
Z
Z
1
1 3
1 w11
2 3
10
2 10 1
dw =
+C =
(t − 3)11 + C
w10 dw =
t (t − 3) dt = t w
2
3t
3
3 11
33
2
14.
Z
x2 (1 + 2x3 )2 dx
1
Let w = 1 + 2x3 , so dw = 6x dx, or dx =
dw
6x
Z
Z
1 w3
1
1
dw =
+C =
(1 + 2x3 )3 + C
x2 (1 + 2x3 )2 dx = x2 w2
6x
6 3
18
15.
Z
x(x2 + 3)2 dx
1
dw
Let w = x2 + 3, so dw = 2x dx, or dx =
2x
Z
Z
1 w3
1
1
dw =
+ C = (x2 + 3)3 + C
x(x2 + 3)2 dx = xw2
2x
2 3
6
16.
Z
x(x2 − 4)7/2 dx
1
Let w = x2 − 4, so dw = 2x dx, or dx =
dw
2x
Z
Z
1 w9/2
1
1
dw =
+ C = (x2 − 4)9/2 + C
x(x2 − 4)7/2 dx = xw7/2
2x
2 9/2
9
17.
Z
y 2 (1 + y)2 dy
Trick (substitution) question: substitution seems not to work well here, because both factors have
y 2 in them, so neither one is the derivative of the other. We’re better off expanding the (1 + y)2
factor and then integrating each term separately:
Z
Z
Z
2
1
1
y 2 (1 + y)2 dy = y 2 (1 + 2y + y 2 )dy = y 2 + 2y 3 + y 4 = y 3 + y 4 + y 5 + C
3
4
5
Z
18.
(2t − 7)73 dt
1
Let w = 2t − 7, so dw = 2 dt, or dt = dw
2
Z
Z
74
w
1
1
1
+C =
(2t − 7)74 + C
(2t − 7)73 dt = w73 dw =
2
2 74
148
19.
Z
1
dy
y+5
Let w = y + 5, so dw = dy, making
Z
Z
1
dy
dy =
dw = ln |w| + C = ln |y + 5| + C
y+5
w
20.
Z
1
√
dx
4−x
Rewrite integral:
Z
(4 − x)−1/2 dx
Z
Let w = 4 − x, so dw = −1 dx, or dx = −dw
Z
w1/2
= −2(4 − x)1/2 + C
(4 − x)−1/2 dx = w−1/2 (−1)dw = −
1/2
3
21.
Z
(x2 + 3)2 dx
Another non-substitution integral: since there is no x term outside the (x2 + 3), it is easier to
expand the square in this case and integrate term by term.
Z
Z
x5
6x3
x5
2
2
(x + 3) dx = x4 + 6x2 + 9 dx =
+
+ 9x + C =
+ 2x3 + 9x + C
5
3
5
Z
3
22.
x2 ex +1 dx
1
Let w = x3 + 1, so dw = 3x2 dx, or dx = 2 dw
3x
Z
Z
3
1
1 3
1
x2 ex +1 dx = x2 ew 2 dw = ew + C = ex +1 + C
3x
3
3
23.
Z
sin θ(cos θ + 5)7 dθ
Let w = cos θ + 5, so dw = − sin θdθ, making
Z
w8
1
sin θ(cos θ + 5)7 dθ = − w7 dw = −
+ C = − (cos θ + 5)8 + C
8
8
Z
24.
Z p
cos(3t) sin(3t) dt
Let w = cos(3t) so dw = −3 sin(3t) dt, or dt =
Z
Z p
cos(3t) sin(3t) dt = w1/2 sin(3t)
25.
Z
−1
dw
3 sin(3t)
−1
−1 w3/2
−2
dt =
+C =
(cos(3t))3/2 + C
3 sin(3t)
3 3/2
9
sin6 θ cos θ dθ
1
Let w = sin(θ) so dw = cos θ dθ, or dθ =
dw
cos θ
Z
Z
w7
1
1
dw =
+ C = sin7 θ + C
(sin θ)6 cos θ dθ = w6 cos θ
cos θ
7
7
26.
Z
sin3 α cos αdα
1
Let w = sin(α) so dw = cos α dα, or dα =
dw
cos α
Z
Z
w4
1
1
dw =
+ C = sin4 α + C
(sin α)3 cos α dα = w3 cos α
cos α
4
4
27.
Z
sin6 (5θ) cos(5θ) dθ
1
Let w = sin(5θ) so dw = 5 cos(5θ) dθ, or dθ =
dw
5 cos(5θ)
Z
Z
1 w7
1
1
dw =
+C =
sin7 (5θ) + C
(sin(5θ))6 cos(5θ) dθ = w6 cos(5θ)
5 cos(5θ)
5 7
35
4
28.
Z
tan(2x) dx
Rewrite integral:
Z
sin(2x)
dx
cos 2x
−1
Let w = cos(2x), so dw = −2 sin(2x) dx, or dx =
dw
2 sin(2x)
Z
Z
Z
sin(2x) −1
−1
−1
sin(2x)
dx =
dw =
ln(|w|) + C
w−1 dw =
cos(2x)
w 2 sin(2x)
2
2
−1
=
ln(| cos(2x)|) + C
2
29.
Z
(ln z)2
dz
z
1
Let w = ln(z), so dw = dz, or dz = z dw
z
Z 2
Z
w
w3
(ln z)3
(ln z)2
dx =
(z dw) =
+C =
+C
z
z
3
3
30.
Z
et + 1
dt
et + t
1
Let w = et + t, so dw = (et + 1) dt, or dt = t
dw
e +1
Z t
Z
Z t
e +1 1
e +1
dt =
dw = w−1 dw = ln(|w|) + C = ln(|et + t|) + C
et + t
w et + 1
31.
Z
y
dy
y2 + 4
1
Let w = y 2 + 4, so dw = 2y dy, or dy =
dw
2y
Z
Z
Z
y 1
1
1
1
y
dy =
dw =
w−1 dw = ln(|w|) + C = ln(|y 2 + 4|) + C
y2 + 4
w 2y
2
2
2
Note that y 2 + 4 is always positive, so we could remove the absolute values if we wished, as they
are redundant in this case.
√
Z
cos( x)
√
dx
32.
x
√
√
1
Let w = x, so dw = x−1/2 dx, or dx = 2 x dw
2
√
Z
Z
Z
√
cos( x)
cos(w) √
√
√ (2 x dw) = 2 cos(w) dw = 2 sin(w) + C = 2 sin( x) + C
dx =
x
x
33.
Z
√
e y
√ dy
y
√
√
1
Let w = y, so dw = y −1/2 dy, or dy = 2 y dw
2
Z √y
Z
Z w
√
e
e
√
√ dy =
√ (2 y dw) = 2 ew dw = 2ew + C = 2e y + C
y
y
5
34.
Z
1 + ex
√
dx
x + ex
1
Let w = x + ex , so dw = (1 + ex ) dx, or dx =
dw
1 + ex
Z
Z
Z
√
1 + ex
1 + ex
1
w1/2
−1/2
√
√
dx
=
dw
=
w
dw
=
+ C = 2 x + ex + C
x
x
w
1
+
e
1/2
x+e
35.
Z
ex
dx
2 + ex
1
Let w = 2 + ex , so dw = ex dx, or dx = x dw
e
Z
Z x
Z
e
1
ex
dx
=
dw
= w−1 dw = ln(|w|) + C = ln(|2 + ex |) + C
2 + ex
w ex
36.
Z
x2
x+1
dx
+ 2x + 19
1
dw
Let w = x2 + 2x + 19, so dw = (2x + 2) dx, or dx =
2(x + 1)
Z
Z
Z
x+1
x+1
1
1
dx
=
dw
=
w−1 dw
x2 + 2x + 19
w
2(x + 1)
2
1
1
= ln(|w|) + C = ln(|x2 + 2x + 19|) + C
2
2
37.
Z
t
dt
1 + 3t2
1
Let w = 1 + 3t2 , so dw = 6t dt, or dt =
dw
Z
Z
Z6t
t
t
1
1
1
1
dt =
dw =
w−1 dw = ln(|w|) + C = ln(|1 + 3t2 |) + C
1 + 3t2
w 6t
6
6
6
In this case we can remove the absolute values because 1 + 3t2 will always be positive, so the
absolute values are redundant.
Z x
e − e−x
dx
38.
ex + e−x
1
Let w = ex + e−x , so dw = (ex − e−x ) dx, or dx = x
dw
e − e−x
Z
Z
Z x
ex − e−x
1
e − e−x
dx =
dw = w−1 dw = ln(|w|) + C = ln(|ex + e−x |) + C
x
−x
x
e +e
w
e − e−x
In this case we can remove the absolute values because ex + e−x will always be positive, so the
absolute values are redundant.
Z
(t + 1)2
dt This question is probably more easily solved by expanding than by using substitu39.
t2
tion.
Z
Z 2
Z
(t + 1)2
t + 2t + 1
2
1
dt =
dt = 1 + + 2 dt = t + 2 ln(|t|) − t−1 + C
t2
t2
t
t
1
= t + 2 ln(|t|) − + C
t
6
40.
Z
x cos(x2 )
p
dx
sin(x2 )
Let w = sin(x2 ), so dw = 2x cos(x2 ) dx or dx =
Z
x cos(x2 )
p
dx =
sin(xw )
Z
x cos(x2 )(w−1/2
1
1
dw =
2x cos(x2 )
2
7
Z
dw
2x cos(x2 )
w−1/2 dw =
p
1 w1/2
+ C = sin(x2 ) + C
2 1/2
8
9
10
11