Solution to Exercise 1:
The problem was to use simple substitution to find the indefinite integral of
|
x e-x2/2 dx
|
Step 1) What part of the integrand is the derivative of another part, and is that derivative
a factor of the integrand? The
x up front is the derivative of the
x2/2 that appears inside the
exponential. That up-front
x also meets the second requirement -- that it
be a factor of the integrand (that is it is multiplied by the rest of the integrand).
So at this point you've identified the two pieces of the integrand you need in
order to proceed.
Step 2) Prepare your substitution variable. In step 1, you identified the
up-front x as being the derivative of the
x2/2. So let
u = x2/2. Then
du
= x
dx
and when you multiply through by
dx you find that
du = x dx.
Step 3) Make the substitution.
Remember the tip about multiplication being commutative. That means that this integral is the
same as
|
e-x2/2 x dx
|
So now
x dx becomes
du and
x2/2 becomes
u. Your
substituted integral is
|
e-u du
|
Step 4) Integrate the substituted integral.
Use the rules from the table and equation
11.2-9 to help you with this. You ought to get
|
e-u du = -e-u + C
|
Step 5) Substitute back.
Your original substitution was u = x2/2.
So put that back into the expression you got by integrating the substituted integral.
|
e-u du = -e-u + C = -e-x2/2 + C
|
That is your answer. Now take the derivative of your answer
(using the
chain rule)
to verify
that you get back the original integrand.
Return to main text
Solution to Exercise 2
The problem was to use simple substitution of variables to find the indefinite
integral of
|
__________
sin(θ) √1 - cos(θ) dθ
|
Step 1) What part of the integrand is the derivative of another part? Clearly
sin(θ) is the derivative of
-cos(θ). But if you substituted
u = -cos(θ), you'd have some
stuff left over under the radical after you made the substitution. Observe that
sin(θ) is
also the
derivative of
1 - cos(θ), and by
substituting
u = 1 - cos(θ) you will
end up with more simplification.
Step 2) Prepare your substitution variable. If we make the substitution,
u = 1 - cos(θ),
then
du
= sin(θ)
dθ
and when you multiply through by
dθ, you get
du = sin(θ) dθ
Now you're ready for
Step 3) Make the substitution.
Remember about the commutativity of multiplication that enables you
to move the cos(θ) so
that it is next to the dθ. When
you do that and make the substitution, you will get
|
_
√u du
|
Step 4) Integrate the substituted integral.
You've seen this particular integral before. Remember that
taking the square root is the same as raising something
to the 1/2 power, and that you can apply equation
11.2-4f to that.
|
_
√u du =
|
2
3
|
u3/2 + C
|
Step 5) Substitute back.
Your original substitution for this was
u = 1 - cos(θ). So substitute
back for u
|
_ 2
√u du =
3
|
u3/2 + C =
|
2
3
|
(1 - cos(θ))3/2 + C
|
which is the answer. Finally, you get to do the step that automatically follows --
that is verifying your answer by matching its derivative with the original
integrand.
Return to main text
Solution to Exersice 3
The problem was to find the indefinite integral of
|
cos3(θ) dθ
|
The hint told you to use the identity,
cos2(θ) = 1 - sin2(θ).
To do that you have to find a
cos2(θ) in the
integral itself. So the first thing to do here is
Step 0) Convert the integrand to a form on which substitution will work.
|
cos3(θ) dθ =
|
|
cos(θ) cos2(θ) dθ =
|
|
cos(θ) (1 - sin2(θ)) dθ
|
The hint also recommended that you should break this into the
difference of two separate integrals.
|
cos(θ) (1 - sin2(θ)) dθ =
|
|
cos(θ) dθ -
|
|
cos(θ) sin2(θ) dθ
|
As predicted in the hint, the first integral in the difference is easy. For the second you need
to
Step 1) Find what part of the integrand is the derivative of another part.
Clearly cos(θ) is
the derivative of
sin(θ). So that will form
the basis of the substitution for the second integral in the difference.
Step 2) Prepare your substitution variable.
If you substitute
u = sin(θ),
then
du
= cos(θ)
dθ
Then multiplying through by
dθ, you get
du = cos(θ) dθ
Now, in the second integrand,
move the
cos(θ)
next to the
dθ (which you can do by commutativity
of multiplication), and you're ready to
Step 3) Make the substitution.
Remember that you are substituting only the right-hand integral of the
difference. Leave the left-hand integral in the difference alone. In
the right-hand integral, substitute
u for sin(θ)
and du for
cos(θ) dθ.
|
cos(θ) dθ -
|
|
sin2(θ) cos(θ) dθ =
|
|
cos(θ) dθ -
|
|
u2 du
|
Step 4) Integrate.
The left-hand integral in the difference is easy and you can find it
on the table. The right-hand integral
succumbs to equation 11.2-4f.
|
cos(θ) dθ -
|
|
u2 du = sin(θ) -
|
1
3
|
u3 + C
|
(Again we combined the two undetermined constants from the two integrals
into a single
C)
Step 5) Substitute back.
Your original substitution was
u = sin(θ). So put
sin(θ) in everywhere
you see a u. You get
which is the answer. Take its derivative and see how you can
manipulate that back into the original integrand.
Return to main text
Solution to Exercise 4
The problem was to use simple substitution to integrate
|
2θ cos(θ2) sin(θ2) dθ
|
Step 1) What part of the integrand is the derivative of another part?
If you had remarkably keen perception, you might have observed that
2θ cos(θ2)
is the derivative of
sin(θ2). But
those who have progressed far enough to see that probably don't need to
be reading this tutorial. Most folks, though,
would see that
2θ is the derivative of
θ2. So we'll
go with that and see where it leads.
Step 2) Prepare your substitution variable.
If you are going to substitute
u = θ2,
then
du
= 2θ
dθ
And when you multiply through by
dθ, you
get
du = 2θ dθ
When you move the
2θ in the
original integrand so that it is proximate to the
dθ,
you find that you are ready to
Step 3) Make the substitution.
So you substitute θ2 with
u and 2θ dθ
with du.
|
cos(θ2) sin(θ2) (2θ) dθ =
|
|
cos(u) sin(u) du
|
The new integrand is simpler than the one you started with but still
presents difficulty. Remember the hint said that sometimes once is
not enough. This new integrand is susceptible to further substitution.
Why? Because
cos(u) is the derivative of
sin(u). So if you let
v = sin, then
dv
= cos(u)
du
and multiplying through by
du you get
dv = cos(u) du
Now substitute again -- this time replace
sin(u)
with
v and
cos(u) du with
dv.
|
cos(θ2) sin(θ2) (2θ) dθ =
|
|
cos(u) sin(u) du =
|
|
v dv
|
Step 4) Integrate the substituted integrand.
This one is pretty easy. Remember that
v = v1,
and apply equation
11.2-4f. You find that
|
1
v dv =
2
|
v2 + C
|
Step 5) Substitute back.
Since you substituted twice, you will have to substitute back twice -- in the
reverse order that you substituted going in. The last substitution you made was
v = sin(u). So that is the first back substitution
to make:
|
1
v dv =
2
|
v2 + C =
|
1
2
|
sin2(u) + C
|
Now back substitute using the first of your substitutions, which was
u = θ2.
|
1
v dv =
2
|
v2 + C =
|
1
2
|
sin2(u) + C =
|
1
2
|
sin2(θ2) + C
|
To verify this answer by taking the derivative, you will have to employ
the
chain rule twice.
PS)
If you did have the keen perception to see that
2θ cos(θ2)
is the derivative of
sin(θ2),
you would have been able to do this integral problem with just a single substitution
instead of having to substitute twice. Now that you know, go ahead and try doing
that way.
Return to main text
Solution to Exercise 5
The problem was to use substitution of variables to integrate
|
(3x2 - 6x + 2) dx
x3 - 3x2 + 2x - 9
|
Step 1) What part of the integrand is the derivative of another part?
In this one the numerator expression (sans the
dx) is the derivative of the
denominator expression. We'll work this one through the remaining steps, and then
I'll have a comment about the general case of when the numerator is the derivative
of the denominator.
Step 2) Prepare your substitution variable.
You will be substituting the entire denominator. So you will have
u = x3 - 3x2 + 2x - 9.
Then
du
= 3x2 - 6x + 2
dx
and multiplying through by
dx, you get
du = (3x2 - 6x + 2) dx
Step 3) Make the substitution.
So you will substitute u = x3 - 3x2 + 2x - 9
with u and
(3x2 - 6x + 2) dx with du.
|
(3x2 - 6x + 2) dx
=
x3 - 3x2 + 2x - 9
|
|
du
u
|
Step 4) Integrate the substituted integral.
If you don't remember how to integrate this, see equation
11.2-6c.
|
(3x2 - 6x + 2) dx
=
x3 - 3x2 + 2x - 9
|
|
du
= ln|u| + C
u
|
Step 5) Substitute back.
Your original substitution was
u = x3 - 3x2 + 2x - 9,
which is what you use to substitute back.
|
(3x2 - 6x + 2) dx
=
x3 - 3x2 + 2x - 9
|
|
du
= ln|u| + C = ln|x3 - 3x2 + 2x - 9| + C
u
|
The expression on the right is your answer. Take its derivative using
the
chain rule to verify it.
Something to remember:
Whenever you have a quotient that you need to integrate, and
the numerator is exactly equal to the derivative of the denominator,
the integral will always be equal to the natural log
of the denominator. That is, if you have
|
f(x) dx
g(x)
|
and
dg
f(x) =
dx
then
|
f(x) dx
= ln|g(x)| + C
g(x)
|
If you take the derivative of
ln|g(x)|, you will see why.
Another way to look at it is this:
|
f(x) dx
=
g(x)
|
|
dg
dx
dx
=
g
|
|
dg
= ln|g| + C
g
|
Remember in the main text when we found the integral of
cot(θ) dθ?
That was an example of exactly this phenomenon. Can you use a slight variation on it
to find
|
tan(θ) dθ
|
Return to main text
Solution to Exercise 6
The problem was to use substitution of variables to integrate
|
______
x √3x - 7 dx
|
Step 1)
Observe that the stuff under the radical is a linear function of x.
That means that if you use that as your substitution function
(i.e.
u = 3x - 7), you will be
able to solve for
x in terms of
u easily.
Step 2) Prepare your substitution variable. If you substitute
u = 3x - 7, then
du
= 3
dx
and
du = 3 dx
To get a
3 dx in the integrand, you will have
to multiply it by
3/3 and pull the denominator outside of the integral:
1
3
|
|
______
x √3x - 7 (3) dx
|
And in order to substitute that
x to the left of the radical
you also need to solve for
x in terms of
u.
u = 3x - 7
so
u + 7
x =
3
Now you are ready to
Step 3) Make the substitution.
You are substituting u for 3x - 7,
and du for 3 du. And where the x
appears by itself, you are going to substitute it with (u + 7)/3.
1
3
|
|
______
x √3x - 7 (3) dx =
|
1
3
|
|
u + 7
3
|
|
_
√u du =
|
1
9
|
|
(u + 7) u1/2 du
|
Step 4) Multiply through and then break it into two the sum of two integrals.
That gives you
1
9
|
|
(u + 7) u1/2 du =
|
1
9
|
|
(u3/2 + 7u1/2) du =
|
1
9
|
|
u3/2 du +
|
7
9
|
|
u1/2 du
|
Step 5) Integrate. Use equation 11.2-4f to
integrate these two summands:
1
9
|
|
u3/2 du +
|
7
9
|
|
u1/2 du =
|
2
45
|
u5/2 +
|
14
27
|
u3/2 + C
|
Step 6) Substitute back.
Your original substitution function was u = 3x - 7.
So
2
45
|
u5/2 +
|
14
27
|
u3/2 + C =
|
2
45
|
(3x - 7)5/2 +
|
14
27
|
(3x - 7)3/2 + C
|
Now take the derivative of this answer and do the appropriate algebraic munging
to verify that you can get back to the original integrand.
Return to main text
Solution to Exercise 7
The problem was to use simple substitution of variables to find the indefinite integral of
|
dx
2(√x + 1 + 7) √x + 1
|
Step 1) What part of the integrand is the derivative of another part? The hint
mentioned that when you take the derivative of a square root you get one half the reciprocal
of the square root. The observation to make on this one is that
1
2√x + 1
is the derivative of
_____
√x + 1 + 7
Step 2) Prepare your variables for substitution.
If you let
_____
u = √x + 1 + 7
then
du 1
=
dx 2√x + 1
and multiplying through by
dx you get
dx
du =
2√x + 1
Step 3) Make the substitution. You have to use the commutative law of addition to
move the 2 in the denominator so that it groups with the last factor in the
denominator, and when you do, the substitution is perfect:
|
dx
=
2(√x + 1 + 7) √x + 1
|
|
du
u
|
Step 4) Integrate the substituted integral.
You've seen this one before. You can use equation
11.2-6c to integrate this.
|
du
= ln|u| + C
u
|
|
Step 5) Substitute back. Your original substitution was
_____
u = √x + 1 + 7
So put that expression in for
u and you get
|
dx
=
2(√x + 1 + 7) √x + 1
|
|
du
= ln|u| + C =
u
|
_____
ln|√x + 1 + 7| + C
|
which is the answer. As always, take the derivative of this using the
chain rule to verify that
you get back the original integrand. When you're done with that, ask yourself,
if this same integral problem had been presented in the equivalent form of
|
dx
2x + 2 + 14√x + 1
|
|
would you have figured out that you had to factor the
_____
2√x + 1
out of the denominator to get it into a form where simple substitution would work?
Look carefully at how that works, and remember that often you will have to do
algebraic manipulations to put an integrand into a form that you know how to
integrate.
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