Example
Z
Evaluate
3x 2
p
4
x 3 + 4 dx.
Solution:
We need a new technique to integrate this function.
Notice that if we let u = x3 + 4, and we compute the differential du of u, we get:
du = 3x2 dx
Going back to our integral, notice that we can eliminate every x that appears in the integral
and write it completely in terms of u using both the definition of u and its differential.
Z
Z 1/4 p
4
3x 2 x 3 + 4 dx =
x3 + 4
· 3x2 dx
Z
=
u1/4 · du
In this process, we took an integral that looked difficult and with a substitution we were
able to rewrite the integral into a simpler looking one that we can do. Evaluating this gives:
Z
Z
p
4
3x 2 x 3 + 4 dx =
u 1/4 du
Using u = x 3 + 4 and du = 3x 2 dx.
=
=
AMAT 217 (University of Calgary)
4u 5/4
+C
5
3
4(x + 4)5/4
+C
5
Using the power rule for integrals.
Converting back to x’s.
Fall 2013
1 / 24
Substitution Rule:
If u = g (x) is a differentiable function whose range is an interval I and f is continuous on I , then
Z
f (g (x))g 0 (x) dx =
Z
f (u) du.
Proof:
This is proved by differentiating the chain rule.
For the function composition F ◦ g , the chain rule states
(F (g (x)))0 = F 0 (g (x))g 0 (x).
Taking antiderivatives (and switching sides) gives:
Z
F 0 (g (x))g 0 (x) dx = F (g (x)) + C .
Now, by setting u = g (x) we get:
Z
Z
F 0 (g (x))g 0 (x) dx = F (u) + C =
F 0 (u) du.
Finally, writing F 0 = f , we have:
Z
Z
f (g (x))g 0 (x) dx =
f (u) du.
AMAT 217 (University of Calgary)
Fall 2013
2 / 24
Notes:
You can think of the Substitution Rule as being the “opposite of the Chain Rule”.
The Substitution Rule allows us to turn a difficult looking integral into one that is easier to
compute.
Tip: Choose a substitution u so that its derivate appears in the integral (up to a constant).
A general strategy to follow is as follows.
Substitution Rule Strategy
1
Choose a possible u = u(x), try one that appears inside a function.
2
Calculate du = u 0 (x) dx.
3
Either replace u 0 (x) dx by du, or replace dx by
4
Write the rest of the integrand in terms of u.
5
Integrate.
6
Rewrite the result back in terms of x.
AMAT 217 (University of Calgary)
du
, and cancel.
u 0 (x)
Fall 2013
3 / 24
Example
Z
Evaluate
x sin(x 2 ) dx
Solution:
Since x 2 appears inside of a function we can try the substitution u = x2 .
Taking the derivative gives:
du
= 2x
dx
→
du = 2x dx
→
du
= x dx
2
Since an x dx appears in the integrand, this substitution should work:
Z
Z
x sin(x 2 ) dx =
sin(x2 ) x dx
Rearranging the integrand
Z
=
=
1
2
=
−
=
AMAT 217 (University of Calgary)
sin(u)
du
2
Using u = x 2 and
du
= x dx
2
Z
sin(u) du
1
cos(u) + C
2
1
− cos(x2 ) + C
2
Pulling the constant out in front
Z
Since integral of
sin x dx = − cos x + C .
Convert to x using u = x2 .
Fall 2013
4 / 24
Example
Z
Evaluate the following integral:
√
cos( x)
dx.
√
x
We use the substitution:
u=
Then the derivative is:
√
x.
1
du = √ dx.
2 x
Solving for dx gives:
√
dx = 2 x du
The integral becomes:
Z
√
cos( x)
dx
√
x
cos u √
2 x du
√
x
Z
=
Z
=
2
=
2 sin u + C
√
2 sin( x) + C
=
AMAT 217 (University of Calgary)
cos u du
Fall 2013
5 / 24
Example
Z
Evaluate:
2x 3
p
x 2 + 1 dx.
Solution:
It makes sense to let: u = x2 + 1. Taking derivatives gives:
du = 2x dx.
→
Making this substitution
Z
Z
pgives:
√
2x 3 x2 + 1 dx =
2x 3 u
dx =
du
2x
du
2x
Z
=
√
x 2 u du
Issue: Our integrals can’t have two variables in them (this is not allowed until Calculus III).
Usually this means we chose our u incorrectly. But in this case:
u = x 2 + 1 → x2 = u − 1.
Z
Z
√
√
(u − 1) u du
x2 u du =
Z
=
u 3/2 − u 1/2 du
By expanding products!
=
=
AMAT 217 (University of Calgary)
2 5/2
2
u
− u 3/2 + C
5
3
2 2
2
(x + 1)5/2 − (x 2 + 1)3/2 + C
5
3
Fall 2013
6 / 24
Example
Z
Evaluate
sin3 x cos x dx
Solution:
We can try u = sin x or u = cos x (note: both will actually work for this example).
Using u = sin x and du = cos x dx, that is, dx =
Z
sin3 x cos x dx
Z
=
Z
AMAT 217 (University of Calgary)
u3 cos x
du
cos x
du
:
cos x
Using the substitution
=
u 3 du
=
u4
+C
4
Since
=
sin4 x
+C
4
Replacing u back in terms of x
Canceling (writing integral in terms of u’s)
Z
x n dx =
x n+1
+ C , n 6= −1
n+1
Fall 2013
7 / 24
Products of Sine and Cosine
We use the following guidelines to determine a suitable substitution when dealing with
products of sine and cosine:
Products of Sine and Cosine
Z
When evaluating
sinm x cosn x dx:
1
The power of sine is odd (m odd):
(a) Use u = cos x and du = − sin x dx.
(b) Cancel one sin x from the dx replacement to be left with an even number of sines.
(c) Use sin2 x = 1 − cos2 x (= 1 − u 2 ) to replace the leftover sines.
2
The power of cosine is odd (n odd):
(a) Use u = sin x and du = cos x dx.
(b) Cancel one cos x from the dx replacement to be left with an even number of cosines.
(c) Use cos2 x = 1 − sin2 x (= 1 − u 2 ) to replace the leftover cosines.
3
The power of both sine and cosine are odd (m and n are odd):
Use either 1 or 2 (both will work).
4
The power of both sine and cosine are even (m and n are even):
Use cos2 x = 12 (1 + cos(2x)) and/or sin2 x = 12 (1 − cos(2x)) to reduce to a form that can
be integrated.
AMAT 217 (University of Calgary)
Fall 2013
8 / 24
Example
Z
Evaluate
sin6 x cos5 x dx
Solution:
Power of cos is odd, therefore use u = sin x and du = cos x dx, that is, dx =
Z
sin6 x cos5 x dx
Z
=
Z
=
AMAT 217 (University of Calgary)
Using the substitution
u 6 cos2 x
2
Canceling a cos x and rewriting cos4 x
du
Using trig identity cos2 x = 1 − sin2 x
Z
u 6 (1 − u 2 )2 du
Writing integral in terms of u’s
u 6 − 2u 8 + u 10 du
Expand and collect like terms
Z
=
du
cos x
u 6 (1 − sin2 x)2 du
=
=
u 6 cos5 x
Z
=
=
du
:
cos x
u7
2u 9
u 11
−
+
+C
7
9
11
7
9
sin x
2 sin x
sin11 x
−
+
+C
7
9
11
Z
Since
x n dx =
x n+1
+ C , n 6= −1
n+1
Replacing u back in terms of x
Fall 2013
9 / 24
Example
Z
Evaluate
cos3 x dx
Solution:
Power of cos is odd, therefore use u = sin x and du = cos x dx, that is, dx =
du
.
cos x
This may seem strange at first since we don’t have sin x in the question, but it does work!
Z
cos3 x dx
Z
=
Z
Using the substitution
Canceling a cos x
Z
(1 − sin2 x) du
Using trig identity cos2 x = 1 − sin2 x
Z
(1 − u 2 ) du
Writing integral in terms of u’s
=
u3
=
u−
=
sin x −
AMAT 217 (University of Calgary)
du
cos x
cos2 x du
=
=
cos3 x
3
Z
x n dx =
x n+1
+ C , n 6= −1
n+1
+C
Since
sin3 x
+C
3
Replacing u back in terms of x
Fall 2013
10 / 24
Example
Z
Evaluate
sin2 x cos2 x dx
Solution:
Power of sine and cosine are both even, thus we use the formulas:
1
1
cos2 x = (1 + cos(2x))
sin2 x = (1 − cos(2x))
2
2
Rewriting the integrand we have:
sin2 x cos2 x
=
=
=
Now we have:
Z
sin2 x cos2 x dx
=
=
=
AMAT 217 (University of Calgary)
1
1
(1 + cos(2x)) (1 − cos(2x))
2
2
1
(1 + cos(2x)) (1 − cos(2x))
4
1
1 − cos2 (2x)
4
Z
1
(1 − cos2 (2x)) dx
4
Z
Z
1
1
1 dx −
cos2 (2x) dx
4
4
Using trig identities
Splitting up integral
???
Fall 2013
11 / 24
Solution (CONTINUED):
Z
sin2 x cos2 x dx
=
x
1
−
4
4
Z
cos2 (2x) dx
For the second integral, the power of sine and cosine are both even.
(Although sine does not appear in the integral, you can think of it as sine to the power of
zero and zero is an even number).
Thus we use formula:
cos2 θ =
Z
sin2 x cos2 x dx
1
(1 + cos(2θ))
2
x
1
−
4
8
Z
=
=
1
x
−
4
8
sin(4x)
x+
+C
4
=
x
sin(4x)
−
+C
8
32
AMAT 217 (University of Calgary)
(1 + cos(4x)) dx
Using the formula with θ = 2x
Z
Use
cos(ax + b) dx =
1
sin(ax + b) + C
a
Since x/4 − x/8 = x/8
Fall 2013
12 / 24
An example with three answers?
Example
Z
Evaluate
sin x cos x dx
Solution 1:
We use the double angle formula sin(2x) = 2 sin x cos x.
Z
Z
sin 2x
1
sin(2x) dx
Replacing sin x cos x by
sin x cos x dx =
2
2
This is a standard substitution with u = 2x and du = 2dx, that is, dx =
Z
sin x cos x dx
AMAT 217 (University of Calgary)
1
2
Z
=
1
4
Z
=
=
1
(− cos u) + C
4
=
−
sin(u)
du
2
sin u du
1
cos(2x) + C
4
du
:
2
Using the substitution
Pulling out the constant
Z
Since
sin x dx = − cos x + C
Replacing u back in terms of x
Fall 2013
13 / 24
An example with three answers?
Example
Z
Evaluate
sin x cos x dx
Solution 2:
We use the substitution u = sin x so that du = cos x dx .
Z
Z
sin x cos x dx
=
u du
Using our substitution
Z
2
x n+1
u
+C
Since
x n dx =
+ C , n 6= −1
=
2
n+1
1
=
sin2 x + C
2
Solution 3:
We use the substitution u = cos x so that du = − sin x dx.
Z
Z
sin x cos x dx = − u du
Using our substitution
Z
2
u
x n+1
= −
+C
Since
x n dx =
+ C, n =
6 −1
2
n+1
1
= − cos2 x + C
2
AMAT 217 (University of Calgary)
Fall 2013
14 / 24
Solution 1 gave an answer of − 41 cos(2x) + C
Solution 2 gave an answer of
1
2
sin2 x + C
Solution 3 gave an answer of − 21 cos2 x + C
Question: Which one is correct? Answer: All of them!
Calculus I: Two functions that have the same derivative differ by a constant.
In each solution the +C is misleading and is actually a different constant for each answer:
1
1
1
Answer 1: − cos(2x)+C1
Answer 2:
sin2 x +C2
Answer 3: − cos2 x +C3
4
2
2
Plotting the three functions without constants confirms that they have the same derivative:
Algebraically you can also show each of the three functions differ by at most a constant.
For example, Answer 2 and Answer 3 are equal when C3 = C2 + 21 :
Answer 2 =
1
1
1
1
1
2
2
2
2
sin x + C2 =
1 − cos x + C2 = − cos x + C2 +
= − cos x + C3 = Answer 3
2
2
2
2
2
AMAT 217 (University of Calgary)
Fall 2013
15 / 24
We next look at integrals with secant and tangent. Some we already know how to do:
Z
Z
sec x tan x dx = sec x + C
sec2 x dx = tan x + C
We can also integrate tan x quite easily.
Example
Z
Evaluate
tan x dx
sin x
Note that tan x =
and let u = cos x,
cos x
Z
Z
sin x
tan x dx =
dx
cos x
Z
du
sin x
=
u
− sin x
Z
1
= −
du
u
so that du = − sin x dx, i.e., dx =
du
:
− sin x
Rewriting tan x
Using the substitution
Cancelling and pulling the −1 out
Z
1
Since
dx = ln |x| + C
x
=
− ln |u| + C
=
− ln | cos x| + C
Replacing u back in terms of x
=
ln | sec x| + C
Using log properties and sec x = 1/ cos x
AMAT 217 (University of Calgary)
Fall 2013
16 / 24
Integrating secant
Example
Z
Evaluate
sec x dx
This is a tough integral.
You won’t need to know how to evaluate this integral for this course.
Weird Trick: We multiply the top and bottom by sec x + tan x and use u = sec x + tan x.
Then, du = (sec x tan x + sec2 x)dx .
Z
Z
sec x dx
=
sec x(sec x + tan x)
dx
sec x + tan x
Z sec2 x + sec x tan x
=
sec x + tan x
Z
=
1
du
u
Doing the trick.
dx
Expanding the top.
Using the substitution.
Z
1
dx = ln |x| + C
x
=
ln |u| + C
Since
=
ln | sec x + tan x| + C
Replacing u back in terms of x
AMAT 217 (University of Calgary)
Fall 2013
17 / 24
Products of Secant and Tangent
We use the following guidelines to determine a suitable substitution when dealing with
products of secant and tangent:
Products of Secant and Tangent
Z
When evaluating
secm x tann x dx:
1
The power of secant is even (m even):
(a) Use u = tan x and du = sec2 x dx.
(b) Cancel sec2 x from the dx replacement to be left with an even number of secants.
(c) Use sec2 x = 1 + tan2 x (= 1 + u 2 ) to replace the leftover secants.
2
The power of tangent is odd (n odd):
(a) Use u = sec x and du = sec x tan x dx.
(b) Cancel one sec x and one tan x from the dx replacement.
The number of remaining tangents is even.
(c) Use tan2 x = sec2 x − 1 (= u 2 − 1) to replace the leftover tangents.
3
m is even or n is odd:
Use either 1 or 2 (both will work).
4
The power of secant is odd and
Z the power of Ztangent is even:
No guidelines. Remember that
sec x dx and
sec3 x dx are in integral tables.
AMAT 217 (University of Calgary)
Fall 2013
18 / 24
Example
Z
Evaluate
sec6 x tan6 x dx
Power of secant is even. Use u = tan x, so that du = sec2 x dx, i.e., dx =
Z
sec6 x tan6 x dx
Z
sec6 x (u 6 )
=
Z
Cancelling
Z
(sec2 x)2 (u 6 ) du
Rewriting sec4 x
Z
(1 + tan2 x)2 (u 6 ) du
Using sec2 x = 1 + tan2 x
Z
(1 + u 2 )2 (u 6 ) du
Using the substitution u = tan x
(u 6 + 2u 8 + u 10 ) du
Expanding.
Z
x n+1
Since
x n dx =
+ C , n 6= −1
n+1
=
=
=
Z
=
=
AMAT 217 (University of Calgary)
Using the substitution
sec4 x(u 6 ) du
=
=
du
sec2 x
du
:
sec2 x
u7
2u 9
u 11
+
+
+C
7
9
11
7
9
tan x
2 tan x
tan11 x
+
+
+C
7
9
11
Replacing u back in terms of x
Fall 2013
19 / 24
Example
Z
Evaluate
sec5 x tan x dx
The power of tangent is odd.
Use u = sec x, so that du = sec x tan x dx, i.e., dx =
du
.
sec x tan x
Remember: Before we replace sec x by u, we need to save one to cancel!!
Z
sec5 x tan x dx
Z
=
Z
=
Z
=
=
=
AMAT 217 (University of Calgary)
sec5 x tan x
du
sec x tan x
Substituting dx first
sec4 x du
Cancelling
u 4 du
Now substituting u = sec x
Z
x n+1
Since
x n dx =
+ C , n 6= −1
n+1
u5
+C
5
5
sec x
+C
5
Replacing u back in terms of x
Fall 2013
20 / 24
Example
Z
Evaluate
sec5 x tan3 x dx
Power of tan is odd. Use u = sec x, so that du = sec x tan x dx, i.e., dx =
Z
Remember: Before we replace sec x by u, we need
Z
du
sec5 x tan3 x dx =
sec5 x tan3 x
sec x tan x
Z
=
sec4 x tan2 x du
Z
=
u 4 tan2 x du
Z
=
u 4 (sec2 x − 1) du
Z
=
u 4 (u 2 − 1) du
Z
=
(u 6 − u 4 ) du
=
=
AMAT 217 (University of Calgary)
u7
u5
−
+C
7
5
7
sec x
sec5 x
−
+C
7
5
du
.
sec x tan x
to save one to cancel!!
Substituting dx first
Cancelling
Now substituting u = sec x
Using tan2 x = sec2 x − 1
Using the substitution
Expanding
Z
x n+1
Since
x n dx =
+ C , n 6= −1
n+1
Replacing u back in terms of x
Fall 2013
21 / 24
Extra Problems:
Example
Z
Evaluate
sec2 x tan6 x dx
The power of secant is even.
We use u = tan x, so that du = sec2 x dx, i.e., dx =
Z
sec2 x tan6 x dx
Z
=
Z
=
=
=
AMAT 217 (University of Calgary)
u7
sec2 x (u 6 )
u 6 du
+C
7
tan7 x
+C
7
du
sec2 x
du
:
sec2 x
Using the substitution
Cancelling
Z
x n+1
Since
x n dx =
+ C , n 6= −1
n+1
Replacing u back in terms of x
Fall 2013
22 / 24
Example
Z
Evaluate
sec4 x tan6 x dx
The power of secant is even.
We use u = tan x, so that du = sec2 x dx, i.e., dx =
Z
sec4 x tan6 x dx
Z
=
Z
Cancelling
(1 + tan2 x)(u 6 ) du
Using sec2 x = 1 + tan2 x
Z
(1 + u 2 )(u 6 ) du
Using the substitution u = tan x
(u 6 + u 8 ) du
Expanding.
Z
x n+1
Since
x n dx =
+ C , n 6= −1
n+1
=
Z
=
AMAT 217 (University of Calgary)
Using the substitution
Z
=
=
du
sec2 x
sec2 x(u 6 ) du
=
=
sec4 x (u 6 )
du
:
sec2 x
u7
u9
+
+C
7
9
7
tan x
tan9 x
+
+C
7
9
Replacing u back in terms of x
Fall 2013
23 / 24
Example
Z
Evaluate
sec x tan2 x dx
The power of secant is odd and the power of tangent is even.
The guidelines don’t help us in this scenario.
We first use the formula: tan2 x = sec2 x − 1
Z
sec x tan2 x dx
Z
=
Z
=
sec x(sec2 x − 1) dx
Trig identity
(sec3 x − sec x) dx
Expanding
=
1
(sec x tan x + ln | sec x + tan x|) − ln | sec x + tan x| + C
2
Formulas
=
1
1
sec x tan x + ln | sec x + tan x| − ln | sec x + tan x| + C
2
2
Expand
=
1
1
sec x tan x − ln | sec x + tan x| + C
2
2
Simplify
AMAT 217 (University of Calgary)
Fall 2013
24 / 24