MAT01A1: Indefinite Integrals and Integration by Substitution · 2020. 5. 23. · On the next slide...

MAT01A1: Indefinite Integrals and Integration bySubstitution

Dr Craig

Week: 25 May 2020

Our previous set of slides covered the

Fundamental Theorem of Calculus (Ch 5.3).

We revise that quickly before moving on to

the two sections covered in these slides: 5.4

and 5.5.

The Fundamental Theorem of Calculus

Suppose f is continuous on [a, b].

1. If g(x) =

∫ xa

f (t) dt, then g′(x) = f (x).

2. ∫ ba

f (x) dx = F (b)− F (a)

where F is a function such that F ′ = f .

We can think of integration and

differentiation as inverse processes.

Warm up

Use FTC2 to calculate the following:

1.

∫ 32

(x2 − 7x + 1) dx

2.

∫ 1−1

(ex + 1) dx

3.

∫ 1−1

(ex +

1

x

)dx

Do these on your own before looking at the

solutions on the slides that follow.

Solutions to warm up exercises

(1.)

∫ 32

(x2 − 7x + 1) dx

=

[x3

3− 7x

2

2+ x

]32

=

(33

3− 7(3

2)

2+ 3

)−(

23

3− 7(2

2)

2+ 2

)=

(9− 63

2+ 3

)−(

8

3− 14 + 2

)=24− 63

2− 8

3=

144− 189− 166

=−61

6

Solutions to warm up exercises

(2.)

∫ 1−1

(ex + 1) dx

= [ex + x]1−1=(e1 + 1

)−(e−1 + (−1)

)=e + 1− 1

e+ 1 = 2 + e + e−1

(3.)

∫ 1−1

(ex +

1

x

)dx We cannot apply

FTC2 to this integral because the integrand

is not continuous on [−1, 1].

Chapter 5.4

Indefinite Integrals


showed us an important relationship between

integration and differentiation. Because of

this relationship, we will use the integral sign

to denote an antiderivative.

∫f (x) dx = F (x)means that F ′(x) = f (x)

For example:∫x2 dx =

x3

3+ C

Indefinite Integrals


showed us an important relationship between

integration and differentiation. Because of

this relationship, we will use the integral sign

to denote an antiderivative.∫f (x) dx = F (x)means that F ′(x) = f (x)

For example:∫x2 dx =

x3

3+ C

Indefinite vs Definite

A definite integral is a number, whereas anindefinite integral is a family of functions.

Remember that the calculation of definite

integrals of the form∫ ba f (x) dx requires that

the function is continuous on [a, b].

When writing down indefinite integrals we

don’t write down the interval on which it is

valid, but this is always implied.

We use the most general antiderivative as the

solution to an indefinite integral (i.e. +C).

On the next slide is an important table of

indefinite integrals.

The first three items in the table are

properties that you have already been using

for definite integrals.

The fourth item is just the power rule of

differentiation in reverse.

The rest of them you also actually know

already. They are simply illustrating the

derivatives that you learned in Chapter 3.

Before we do some examples, let us highlight

a few important indefinite integrals from the

previous slide:

∫1

xdx = ln |x| + C

NB: remember the absolute value!

Two others that are often forgotten are∫1

1 + x2dx = tan−1 x + C

and ∫1√

1− x2dx = sin−1 x + C

Before we do some examples, let us highlight

a few important indefinite integrals from the

previous slide:∫1

xdx = ln |x| + C

NB: remember the absolute value!

Two others that are often forgotten are∫1

1 + x2dx = tan−1 x + C

and ∫1√

1− x2dx = sin−1 x + C

Examples of indefinite integrals:

I∫

(10x4 − 2 sec2 x) dx

I∫

cos θ

sin2 θdθ

I∫

2t2 + t2√t− 1

t2dt

I∫

2s ds

I∫

4√4xdx

Solutions:

I∫

(10x4 − 2 sec2 x) dx

= 2x5 − 2 tanx + C

I∫

cos θ

sin2 θdθ =

∫cos θ

sin θ· 1

sin θdθ

=

∫cot θ csc θ dθ = − csc θ + C

I∫

2t2 + t2√t− 1

t2dt =

∫ (2 +√t− t−2

)dt

= 2t +2t3/2

3+ t−1 + C

Solutions:

I∫

(10x4 − 2 sec2 x) dx = 2x5 − 2 tanx + C

I∫

cos θ

sin2 θdθ =

∫cos θ

sin θ· 1

sin θdθ

=


I∫

2t2 + t2√t− 1

t2dt =

∫ (2 +√t− t−2

)dt

= 2t +2t3/2

3+ t−1 + C

Solutions:

I∫

(10x4 − 2 sec2 x) dx = 2x5 − 2 tanx + C

I∫

cos θ

sin2 θdθ =

∫cos θ

sin θ· 1

sin θdθ

=

∫cot θ csc θ dθ

= − csc θ + C

I∫

2t2 + t2√t− 1

t2dt =

∫ (2 +√t− t−2

)dt

= 2t +2t3/2

3+ t−1 + C

Solutions:

I∫

(10x4 − 2 sec2 x) dx = 2x5 − 2 tanx + C

I∫

cos θ

sin2 θdθ =

∫cos θ

sin θ· 1

sin θdθ

=


I∫

2t2 + t2√t− 1

t2dt =

∫ (2 +√t− t−2

)dt

= 2t +2t3/2

3+ t−1 + C

Solutions:

I∫

(10x4 − 2 sec2 x) dx = 2x5 − 2 tanx + C

I∫

cos θ

sin2 θdθ =

∫cos θ

sin θ· 1

sin θdθ

=


I∫

2t2 + t2√t− 1

t2dt

=

∫ (2 +√t− t−2

)dt

= 2t +2t3/2

3+ t−1 + C

Solutions:

I∫

(10x4 − 2 sec2 x) dx = 2x5 − 2 tanx + C

I∫

cos θ

sin2 θdθ =

∫cos θ

sin θ· 1

sin θdθ

=


I∫

2t2 + t2√t− 1

t2dt =

∫ (2 +√t− t−2

)dt

= 2t +2t3/2

3+ t−1 + C

Solutions continued. . .

First recall thatd

dx(ax) = ax ln a.

I∫

2s ds =2s

ln 2+ C

I∫

4√4xdx =

∫4

2√xdx = 2

∫x−1/2 dx

= 2

(x1/2

1/2

)+C

= 4√x + C


First recall thatd

dx(ax) = ax ln a.

I∫

2s ds

=2s

ln 2+ C

I∫

4√4xdx =

∫4

2√xdx = 2

∫x−1/2 dx

= 2

(x1/2

1/2

)+C

= 4√x + C


First recall thatd

dx(ax) = ax ln a.

I∫

2s ds =2s

ln 2+ C

I∫

4√4xdx =

∫4

2√xdx = 2

∫x−1/2 dx

= 2

(x1/2

1/2

)+C

= 4√x + C


First recall thatd

dx(ax) = ax ln a.

I∫

2s ds =2s

ln 2+ C

I∫

4√4xdx

=

∫4

2√xdx = 2

∫x−1/2 dx

= 2

(x1/2

1/2

)+C

= 4√x + C


First recall thatd

dx(ax) = ax ln a.

I∫

2s ds =2s

ln 2+ C

I∫

4√4xdx =

∫4

2√xdx = 2

∫x−1/2 dx

= 2

(x1/2

1/2

)+C

= 4√x + C

Chapter 5.5

This section is “Integration by substitution”.

You will also often see this technique referred

to as u-substitution. The reason for this

name will soon become clear.

Substitution is used to solve both definite

and indefinite integrals. It is the perfect

place to end this semester’s mathematics.

To find ∫2x√

1 + x2 dx

we use substitution.

We let

u = 1 + x2.

Nowdu

dx= 2x.

At this point, we will treat du and dx like

algebraic terms. (These terms, ∆u and ∆x

are known as differentials, see Ch 3.10 page

254.)

To find ∫2x√

1 + x2 dx

we use substitution. We let

u = 1 + x2.

Nowdu

dx= 2x.

At this point, we will treat du and dx like

algebraic terms. (These terms, ∆u and ∆x

are known as differentials, see Ch 3.10 page

254.)

To find ∫2x√

1 + x2 dx


u = 1 + x2.

Nowdu

dx= 2x

and so (2x)dx = du.

The original integral then becomes∫2x√

1 + x2 dx =

∫ √u du

To find ∫2x√

1 + x2 dx


u = 1 + x2.

Nowdu

dx= 2x and so (2x)dx = du.

The original integral then becomes∫2x√

1 + x2 dx =

∫ √u du

Now we solve the simpler integral:∫ √u du

=

∫u1/2 du =

2

3u3/2 + C

We must then put the integral back to an

expression in terms of x using the fact that

we let u = 1 + x2. We have:∫2x√

1 + x2dx =

∫ √u du

=2

3u3/2+C =

2(1 + x2)3/2

3+C

Now we solve the simpler integral:∫ √u du =

∫u1/2 du

=2

3u3/2 + C




1 + x2dx =

∫ √u du

=2

3u3/2+C =

2(1 + x2)3/2

3+C


∫u1/2 du =

2

3u3/2 + C




1 + x2dx =

∫ √u du

=2

3u3/2+C =

2(1 + x2)3/2

3+C


∫u1/2 du =

2

3u3/2 + C



we let u = 1 + x2.

We have:∫2x√

1 + x2dx =

∫ √u du

=2

3u3/2+C =

2(1 + x2)3/2

3+C


∫u1/2 du =

2

3u3/2 + C




1 + x2dx =

∫ √u du

=2

3u3/2+C =

2(1 + x2)3/2

3+C

In the last example it was relatively easy to

see that we could replace 2x dx with du.

However, it is sometimes easier to solve for

dx and then cancel the terms involving x.

Here is an example of how that would work.

From u = 1 + x2 we get dx =du

2x. This

gives us:

∫2x√

1 + x2dx =

∫2x√udu

2x=

∫ √u du







2x.

This

gives us:

∫2x√

1 + x2dx =

∫2x√udu

2x=

∫ √u du







2x. This

gives us:

∫2x√

1 + x2dx =

∫2x√udu

2x=

∫ √u du

Below is a formal statement of the

substitution rule.

If u = g(x) is a differentiable function

whose range is an interval I and f is

continuous on I , then∫f (g(x))g′(x) dx =

∫f (u) du

Most students find it easier to understand

this rule by looking at examples.

Examples:

1. Find

∫cos3 θ sin θ dθ by letting

u = cos θ.

2. Find

∫x sin(x2) dx

3. Find

∫(x + 1)

√2x + x2 dx.

For the first one we have suggested what

substitution to make. On the next slide we

give a hint for how to find the term to

substitute.

2. Find

∫x sin(x2) dx

3. Find

∫(x + 1)

√2x + x2 dx.

In the two examples above, you have a

composite function multiplied by a term that

looks like the derivative of the inner function

from the composite function.

In cases like this, a good first try for the

substitution is to let u equal to the inner

function of the composite function.

1. Find

∫cos3 θ sin θ dθ by letting u=cos θ.

Solution:

Let u = cos θ. Thendu

dθ= − sin θ,

so −du = sin θdθ. Therefore∫cos3 θ sin θ dθ =

∫u3(−1) du

= −∫u3 du

= −u4

4+ C

= −cos4 θ

4+ C

1. Find


Solution: Let u = cos θ. Thendu

dθ= − sin θ,


∫u3(−1) du

= −∫u3 du

= −u4

4+ C

= −cos4 θ

4+ C

1. Find



dθ= − sin θ,

so −du = sin θdθ. Therefore

∫cos3 θ sin θ dθ =

∫u3(−1) du

= −∫u3 du

= −u4

4+ C

= −cos4 θ

4+ C

1. Find



dθ= − sin θ,


∫u3(−1) du

= −∫u3 du

= −u4

4+ C

= −cos4 θ

4+ C

(2.) Find

∫x sin(x2) dx

Solution:

We let u = x2. Then dudx = 2x, sodu2 = x dx. Therefore∫

x sin(x2) dx =

∫sin(u)

1

2du

=1

2

∫sinu du

=1

2(− cosu) + C

=− cos(x2)

2+ C

(2.) Find

∫x sin(x2) dx

Solution: We let u = x2. Then dudx = 2x,

sodu2 = x dx. Therefore∫

x sin(x2) dx =

∫sin(u)

1

2du

=1

2

∫sinu du

=1

2(− cosu) + C

=− cos(x2)

2+ C

(2.) Find

∫x sin(x2) dx

Solution: We let u = x2. Then dudx = 2x, sodu2 = x dx. Therefore

∫x sin(x2) dx =

∫sin(u)

1

2du

=1

2

∫sinu du

=1

2(− cosu) + C

=− cos(x2)

2+ C

(2.) Find

∫x sin(x2) dx

Solution: We let u = x2. Then dudx = 2x, sodu2 = x dx. Therefore∫

x sin(x2) dx =

∫sin(u)

1

2du

=1

2

∫sinu du

=1

2(− cosu) + C

=− cos(x2)

2+ C

(3.) Find

∫(x + 1)

√2x + x2 dx.

Solution:

Let u = 2x + x2. Thendudx = 2 + 2x. So dx =

du2+2x. We get∫

(x + 1)√

2x + x2 dx =

∫(x + 1)

√u

2(1 + x)du

=1

2

∫ √u du

=1

2

(2

3u3/2

)+ C

=(2x + x2)3/2

3+ C

(3.) Find

∫(x + 1)

√2x + x2 dx.

Solution: Let u = 2x + x2. Thendudx = 2 + 2x.

So dx = du2+2x. We get∫(x + 1)

√2x + x2 dx =

∫(x + 1)

√u

2(1 + x)du

=1

2

∫ √u du

=1

2

(2

3u3/2

)+ C

=(2x + x2)3/2

3+ C

(3.) Find

∫(x + 1)

√2x + x2 dx.

Solution: Let u = 2x + x2. Thendudx = 2 + 2x. So dx =

du2+2x. We get

∫(x + 1)

√2x + x2 dx =

∫(x + 1)

√u

2(1 + x)du

=1

2

∫ √u du

=1

2

(2

3u3/2

)+ C

=(2x + x2)3/2

3+ C

(3.) Find

∫(x + 1)

√2x + x2 dx.

Solution: Let u = 2x + x2. Thendudx = 2 + 2x. So dx =

du2+2x. We get∫

(x + 1)√

2x + x2 dx =

∫(x + 1)

√u

2(1 + x)du

=1

2

∫ √u du

=1

2

(2

3u3/2

)+ C

=(2x + x2)3/2

3+ C

Now is a good point at which to take a

break.

In the last subsection, we will look at:

I some different types of integrals that canbe solved using substitution;

I definite integrals solved by substitution;I symmetry in integration.

Another type of example:∫x5√

1 + x2 dx

This is quite different from the examples that

we have looked at so far. Here we don’t have

a term that looks like the derivative of the

inner function of the composite function.

In order to simplify the integrand, we will

have to do some extra work after making the

initial substitution.

∫x5√

1 + x2 dx

Solution: Let u = 1 + x2.

Then dudx = 2x so

dx = du2x. We get∫x5√

1 + x2 dx =

∫x5√u

2xdu =

1

2

∫x4√u du

But this is no good. We can’t have a mix of

x and u variables. However, we can use our

original substitution to get rid of the x4 term:

u=1 + x2 ⇒ u− 1=x2 ⇒ (u− 1)2=x4

∫x5√

1 + x2 dx

Solution: Let u = 1 + x2. Then dudx = 2x so

dx = du2x.

We get∫x5√

1 + x2 dx =

∫x5√u

2xdu =

1

2

∫x4√u du




u=1 + x2 ⇒ u− 1=x2 ⇒ (u− 1)2=x4

∫x5√

1 + x2 dx



1 + x2 dx =

∫x5√u

2xdu =

1

2

∫x4√u du




u=1 + x2 ⇒ u− 1=x2 ⇒ (u− 1)2=x4

∫x5√

1 + x2 dx



1 + x2 dx =

∫x5√u

2xdu =

1

2

∫x4√u du


x and u variables.

However, we can use our


u=1 + x2 ⇒ u− 1=x2 ⇒ (u− 1)2=x4

∫x5√

1 + x2 dx



1 + x2 dx =

∫x5√u

2xdu =

1

2

∫x4√u du




u=1 + x2 ⇒ u− 1=x2 ⇒ (u− 1)2=x4

∫x5√

1 + x2 dx



1 + x2 dx =

∫x5√u

2xdu =

1

2

∫x4√u du




u=1 + x2

⇒ u− 1=x2 ⇒ (u− 1)2=x4

∫x5√

1 + x2 dx



1 + x2 dx =

∫x5√u

2xdu =

1

2

∫x4√u du




u=1 + x2 ⇒ u− 1=x2

⇒ (u− 1)2=x4

∫x5√

1 + x2 dx



1 + x2 dx =

∫x5√u

2xdu =

1

2

∫x4√u du




u=1 + x2 ⇒ u− 1=x2 ⇒ (u− 1)2=x4

We have (u− 1)2 = x4 and hence we get1

2

∫x4√u du =

1

2

∫(u− 1)2

√u du

Now we multiply out and then integrate:

=1

2

∫(u2 − 2u + 1)

√u du

=1

2

∫ (u5/2 − 2u3/2 + u1/2

)du

=1

2

(2

7u7/2 − 2 · 2

5u5/2 +

2

3u3/2

)+ C

=u7/2

7− 2u

5/2

5+u3/2

3+ C

We started by letting u = 1 + x2 and got:∫x5√

1 + x2 dx =1

2

∫(u− 1)2

√u du

=u7/2

7− 2u

5/2

5+u3/2

3+ C

Therefore we have∫x5√

1 + x2 dx

=(1 + x2)7/2

7− 2(1 + x

2)5/2

5+

(1 + x2)3/2

3+C

Integral of tanx∫tanx dx

Do we know anything that can be

differentiated to give tanx?

No, but we can rewrite this integral as∫sinx

cosxdx

∫tanx dx =

∫sinx

cosxdx

Let u = cosx. Thendu

dx= − sinx which

gives us dx =du

− sinx. Hence

∫tanx dx =∫

sinx

cosxdx =

∫sinx

u· du

(− sinx)

= −∫

1

udu = − ln |u| + C

= − ln | cosx| + C= ln | secx| + C

∫tanx dx =

∫sinx

cosxdx

Let u = cosx.

Thendu

dx= − sinx which

gives us dx =du

− sinx. Hence

∫tanx dx =∫

sinx

cosxdx =

∫sinx

u· du

(− sinx)

= −∫

1

udu = − ln |u| + C


∫tanx dx =

∫sinx

cosxdx


dx= − sinx which

gives us dx =du

− sinx.

Hence∫

tanx dx =∫sinx

cosxdx =

∫sinx

u· du

(− sinx)

= −∫

1

udu = − ln |u| + C


∫tanx dx =

∫sinx

cosxdx


dx= − sinx which

gives us dx =du

− sinx. Hence

∫tanx dx =∫

sinx

cosxdx =

∫sinx

u· du

(− sinx)

= −∫

1

udu = − ln |u| + C


Substitution rule for definite integrals

When we use u-substitution for definite

integrals, we will need to change the limits of

the integral.

Example: Evaluate

∫ 40

√2x + 1 dx.

We let u = 2x + 1. Nowdu

dx= 2 and so

dx = 12du. Now we change the limits.

If x = 0, then u = 1, and if x = 4 we have

u = 9. Thus∫ 40

√2x + 1 dx =

∫ 91

1

2

√u du =

[u3/2

3

]91

=26

3

Example: Evaluate

∫ 40

√2x + 1 dx.

We let u = 2x + 1.

Nowdu

dx= 2 and so



u = 9. Thus∫ 40

√2x + 1 dx =

∫ 91

1

2

√u du =

[u3/2

3

]91

=26

3

Example: Evaluate

∫ 40

√2x + 1 dx.


dx= 2 and so

dx = 12du.

Now we change the limits.


u = 9. Thus∫ 40

√2x + 1 dx =

∫ 91

1

2

√u du =

[u3/2

3

]91

=26

3

Example: Evaluate

∫ 40

√2x + 1 dx.


dx= 2 and so



u = 9. Thus∫ 40

√2x + 1 dx =

∫ 91

1

2

√u du =

[u3/2

3

]91

=26

3

Example: Evaluate

∫ 40

√2x + 1 dx.


dx= 2 and so



u = 9.

Thus∫ 40

√2x + 1 dx =

∫ 91

1

2

√u du =

[u3/2

3

]91

=26

3

Example: Evaluate

∫ 40

√2x + 1 dx.


dx= 2 and so



u = 9. Thus∫ 40

√2x + 1 dx

=

∫ 91

1

2

√u du =

[u3/2

3

]91

=26

3

Example: Evaluate

∫ 40

√2x + 1 dx.


dx= 2 and so



u = 9. Thus∫ 40

√2x + 1 dx =

∫ 91

1

2

√u du

=

[u3/2

3

]91

=26

3

Example: Evaluate

∫ 40

√2x + 1 dx.


dx= 2 and so



u = 9. Thus∫ 40

√2x + 1 dx =

∫ 91

1

2

√u du =

[u3/2

3

]91

=26

3

Notice that when we wrote du, our limits

were in terms of u:∫ 40

√2x + 1 dx =

∫ 91

1

2

√u du =

[u3/2

3

]91

=26

3

Another method that can be used is to

specify that you are leaving the limits in

terms of x, find the integral, and then put x

back in to the indefinite integral that you

have calculated in terms of u.

We show this approach on the next slide.

∫ 40

√2x + 1 dx =

∫ x=4x=0

1

2

√u du

=

[u3/2

3

]x=4x=0

=

[(2x + 1)3/2

3

]40

=(2(4) + 1)3/2

3− (2(0) + 1)

3/2

3

=33

3− 1

3=

26

3

Example:

∫ 1−1

x2

(4− x3)2dx

Let u = 4− x3. Try this one on your own.

Remember, you need to calculate dudx and to

change the limits to u values.

Make sure that you put the correct values for

u in the upper and lower limits. Recall that

we must always have the smaller value as the

lower limit. We can swap limits by

multiplying the integral by −1.

Example:

∫ 1−1

x2

(4− x3)2dx

We let u = 4− x3, so dudx = −3x2. When

x = −1 we get u = 5 and when x = 1 weget u = 3. Therefore∫ 1−1

x2

(4− x3)2dx =

−13

∫ 35

du

u2=

1

3

∫ 53

du

u2

Now

1

3

∫ 53

u−2 du =1

3

[−u−1

]53

=1

3

(−15− −1

3

)=

2

45

Example:

∫ 1−1

x2

(4− x3)2dx

We let u = 4− x3, so dudx = −3x2.

When


x2

(4− x3)2dx =

−13

∫ 35

du

u2=

1

3

∫ 53

du

u2

Now

1

3

∫ 53

u−2 du =1

3

[−u−1

]53

=1

3

(−15− −1

3

)=

2

45

Example:

∫ 1−1

x2

(4− x3)2dx


x = −1 we get u = 5 and when x = 1 weget u = 3.

Therefore∫ 1−1

x2

(4− x3)2dx =

−13

∫ 35

du

u2=

1

3

∫ 53

du

u2

Now

1

3

∫ 53

u−2 du =1

3

[−u−1

]53

=1

3

(−15− −1

3

)=

2

45

Example:

∫ 1−1

x2

(4− x3)2dx



x2

(4− x3)2dx =

−13

∫ 35

du

u2=

1

3

∫ 53

du

u2

Now

1

3

∫ 53

u−2 du =1

3

[−u−1

]53

=1

3

(−15− −1

3

)=

2

45

We have just one small topic from Ch 5.5

still to cover.

Using the symmetry that we learned in

Chapter 1.1, we can make our calculations of

integrals much easier.

Integrals of even functions become much

simpler to calculate when we multiply by 2

and change the bottom limit to 0.

Integrals of Symmetric Functions

Suppose that f is continuous on [−a, a].Then

1. If f is even (i.e. f (−x) = f (x)), then∫ a−af (x) dx = 2

∫ a0

f (x) dx

2. If f is odd (i.e. f (−x) = −f (x)), then∫ a−af (x) dx = 0

Examples of symmetry:

Recall from Ch 1.1 that f (x) = x2 is an even

function and g(x) = x3 is an odd function.∫ 2−2x2dx =

(x3

3

]2−2

=

(8

3

)−(−83

)=

16

3

2

∫ 20

x2 dx = 2

(x3

3

]20

= 2(8

3

)=

16

3∫ 1−1x3 dx =

(x4

4

]1−1

=1

4− 1

4= 0



function and g(x) = x3 is an odd function.

∫ 2−2x2dx =

(x3

3

]2−2

=

(8

3

)−(−83

)=

16

3

2

∫ 20

x2 dx = 2

(x3

3

]20

= 2(8

3

)=

16

3∫ 1−1x3 dx =

(x4

4

]1−1

=1

4− 1

4= 0




(x3

3

]2−2

=

(8

3

)−(−83

)=

16

3

2

∫ 20

x2 dx = 2

(x3

3

]20

= 2(8

3

)=

16

3∫ 1−1x3 dx =

(x4

4

]1−1

=1

4− 1

4= 0




(x3

3

]2−2

=

(8

3

)−(−83

)=

16

3

2

∫ 20

x2 dx = 2

(x3

3

]20

= 2(8

3

)=

16

3

∫ 1−1x3 dx =

(x4

4

]1−1

=1

4− 1

4= 0




(x3

3

]2−2

=

(8

3

)−(−83

)=

16

3

2

∫ 20

x2 dx = 2

(x3

3

]20

= 2(8

3

)=

16

3∫ 1−1x3 dx =

(x4

4

]1−1

=1

4− 1

4= 0

Mixed integration examples:

I∫ex√

1 + ex dx

I∫

2t

2t + 3dt

I∫ e4e

dy

y√`n y

I∫ π/4−π/4

(x3 + x4 tanx) dx

Try all of them on your own first. You can

check your answers to the first two by

differentiating. Solutions follow . . .

Solutions

I∫ex√

1 + ex dx

Let u = 1 + ex. Then dudx = 0 + ex = ex.

So, ex dx = du and hence∫ex√

1 + ex dx =

∫ √u du =

2

3u3/2+C

=2(1 + ex)3/2

3+C


I∫

2t

2t + 3dt Let u = 2t + 3. We get

du

dt= 2t ln 2 =⇒ du

ln 2= 2t dt.

Therefore∫2t

2t + 3dt =

1

ln 2

∫1

udu =

1

ln 2·ln |u|+C

We put t back in. Since 2t + 3 > 0 we get

ln |u|ln 2

+C =ln |2t + 3|

ln 2+C =

ln(2t + 3)

ln 2+C


If you were unsure what to do with the next

example, try re-writing it as follows:∫ e4e

(1

y· 1√

`n y

)dy

I Let u = `n y. Then dudy =1y so du =

1ydy.

Now we change the bounds. When y = e

we get u = 1 and when y = e4, then

u = `n(e4) = 4`n(e) = 4.




(1

y· 1√

`n y

)dy


1ydy.

Now we change the bounds.

When y = e


u = `n(e4) = 4`n(e) = 4.




(1

y· 1√

`n y

)dy


1ydy.

Now we change the bounds. When y = e


u = `n(e4) = 4`n(e) = 4.

Solutions continued. . .∫ e4e

(1

y· 1√

`n y

)dy =

∫ 41

1√udu

=

∫ 41

u−1/2 du

=

[u1/2

1/2

]41

=[2√u]41

= 2√

4− 2√

1

= 2

I∫ π/4−π/4

(x3 + x4 tanx) dx

If we let f (x) = x3 + x4 tanx, then

f (−x) = (−x)3 + (−x)4 tan(−x)= −x3 + x4(− tanx)= −(x3 + x4 tanx)= −f (x).

Since the integrand is an odd function we

have

∫ π/4−π/4

(x3 + x4 tanx) dx = 0.

Khan Academy links

u-substitution

Defininite integrals with u-substitution

https://www.khanacademy.org/math/ap-calculus-ab/ab-integration-new/ab-6-9/v/u-substitutionhttps://www.khanacademy.org/math/ap-calculus-ab/ab-integration-new/ab-6-9/v/u-substitution-definite-integrals

Prescribed tut questions

Ch 5.4:

1, 3, 7, 9, 17, 25, 31, 33, 41

Ch 5.5:

3, 5, 7, 17, 21, 23, 35, 47, 57, 61, 63, 67

MAT01A1: Indefinite Integrals and Integration by Substitution · 2020. 5. 23. · On the next slide...

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