Practice problems: Antidifferentiation by substitution
Bro. David E. Brown, BYU–Idaho Dept. of Mathematics. All rights reserved.
Version 1.0, of May 9, 2013
1
Reminder
The point of antidifferentiating by substitution is to turn the integration problem at hand intoR the antidifferentiation
0 of something whose antiderivative Ris familiar to you. You do this
R
R by thinking of g(x)dx as
f u(x)
u
(x)dx,
which
you
typically
write
as
f
(u)du.
The
hope
is
that
f (u)du is easier to integrate
R
than g(x)dx.
R
Example: Let’s try to antidifferentiate esin x cos x dx. Our g(x) dx is esin x cos x dx. Notice that sin x
is “inside” the exponential function: esin x . Maybe f (u) should be eu and u(x) should be sin x. If so, then
f (u(x)) = esin x , which is part of our integrand; so far, so good.
The rest of our integrand is cos x dx, and cos x is the derivative of sin x. So maybe u0 (x) = cos x. If so,
then we need u(x) to be sin x, which checks out nicely. Evidently,
Z
Z
Z
sin x
u(x) 0
e
cos x dx = e
u (x) dx = eu du.
R
R
R
The last thing to check is whether it’s easier to calculate eu du or esin x cos x dx. No contest!, right? eu du
is definitely the easier of the two. So, we finish by carrying out the antidifferentiation and back-substituting:
Z
Z
sin x
e
cos x dx = eu du = eu + C = esin x + C.
Done.
2
Part A: Exercises on choosing u(x)
Each of the following can be antidifferentiated by substitution. Write down the function you think u(x)
should be, and calculate du. Then determine whether your choice of u actually makes the antidifferentiation
easier to do. Don’t actually carry out the antidifferentiation (unless you actually want to).
R
Exercise 2.1. (x + 1)−2 dx
R√
Exercise 2.2.
2x − 5 dx
R
x
Exercise 2.3. √
dx
2
x +3
R ex
Exercise 2.4.
dx
ex + 1
R
Exercise 2.5. sin3 x cos x dx
R
Exercise 2.6.
cos(sin x) cos x dx (That’s cos x times cos(sin x).)
R ln x
dx
x
R √
Exercise 2.8. x x − 3 dx (Look this one up in the Comments section if you have trouble with it.)
Exercise 2.7.
1
3
Part B: Antidifferentiation by substitution
Calculate each of the following antiderivatives.
R√
2x − 5 dx
Exercise 3.1.
Exercise 3.2.
R
ex
dx
ex + 1
R ln x
dx
x
R
Exercise 3.4. (4x − 7)9 dx
R
2
Exercise 3.5. (2x + 3)ex +3x+1 dx (Yes, all of x2 + 3x + 1 is in the exponent.)
Exercise 3.3.
dx
(x − 1)3
√
R
Exercise 3.7. (r − 1) r2 − 2rdr
R
Exercise 3.8. θ cos(θ2 )dθ (That’s θ times cos(θ2 ).)
R √
Exercise 3.9. x x − 3 dx (See the Comments section.)
R √
Exercise 3.10. x2 x − 3 dx (See the Comments section.)
R
Exercise 3.11. sec2 x tan x dx. Use u(x) = sec x.
R
Exercise 3.12. sec2 x tan x dx. Use u(x) = tan x. Compare with your answer for Exercise 3.11. See the
Comments section for an explanation.)
Exercise 3.6.
4
R
Answers, etc.
Part A
Exercise
Exercise
Exercise
Exercise
Exercise
Exercise
Exercise
Exercise
2.1.
2.2.
2.3.
2.4.
2.5.
2.6.
2.7.
2.8.
u = x + 1, du = dx
u = 2x − 5, du = 2 dx
u = x2 + 3, du = 2x dx
u = ex + 1, du = ex dx
u = sin x, du = cos x dx
u = sin x, du = cos x dx
u = ln x, du = dx
x
u = x − 3, du = dx
Part B
Exercise 3.1.
Exercise 3.2.
Exercise 3.3.
Exercise 3.4.
Exercise 3.5.
Exercise 3.6.
1
(2x − 5)3/2 + C
3 x
ln(e + 1) + C
1 2
1
ln x + C, which is the same as (ln x)2 + C
2
2
1
10
(4x
−
7)
+
C
40
2
ex +3x+1 + C
1
+C
−
2(x − 1)2
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1 2
(r − 2r)3/2 + C
3
1
3.8. sin(θ2 ) + C
2
2
3.9. (x − 3)5/2 + 2(x − 3)3/2 + C
5
2
12
3.10. (x − 3)7/2 + (x − 3)5/2 + 6(x − 3)3/2 + C
7
5
sec2 x
3.11.
+C
2
2
tan x
+C
3.12.
2
Exercise 3.7.
Exercise
Exercise
Exercise
Exercise
Exercise
5
Comments
1. In Part
R A,
√ item (2.8), you really can use u(x) = x − 3. You’ll get du = dx, so at this point you
have
x
R
√udu. Fix that loose x by solving u = x − 3 for x. That is, x = u + 3. So your integral is
(u + 3) udu. I suggest converting the radical to the 1/2 power and multiplying out the integrand.
2. In Part B, items (3.9) and (3.10), proceed as in Part A, item (2.8). In item (3.10), use x = u + 3 to
get x2 = (u + 3)2 , multiply out the integrand, and go for it.
3. For Part B, items (3.11) and (3.12), you get two different-looking answers, namely 12 sec2 x + C1 and
1
2
2 tan x + C2 . (I used C1 in one integral and C2 in the other to be able to distinguish them a few lines
down from here on the page.) These are actually the same answer:
tan2 x + sec2 x = 1,
so
tan2 x = sec2 x − 1.
So 12 tan2 x + C2 = 12 sec2 x + C2 − 1. Since C2 is any old number, C2 − 1 is also any old number. But
then, so is C1 . So we might as well say C1 = C2 − 1, which implies 12 tan2 x + C2 = 21 sec2 x + C2 − 1 =
1
1
1
2
2
2
2 sec x + C1 . So 2 sec x + C1 and 2 tan x + C2 are really the same answer. Weird!
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