EVALUATING INTEGRALS - SUBStitutiON EVALUATING INTEGRALS USING THE SUBSTITUTION RULE (PART I) In previous section, we've dealt with definite and indefinite integrals that can be solved by applying the Fundamental Theorems. However, there's a limit; there are more complex integrals which can't be solved directly with the Fundamental Theorem. Take a good look at this indefinite integral:

∫

4 (1 + 2x)3

dx

Do you think you can solve this integral directly using FTC? Obviously not! You'll realize that FTC isn't exactly well “equipped” to handle relatively complex integrals like the one above (at least, not in this form). So how do we go about it? First, understand the problem: in this form, the integral is too complex for the FTC to handle. So, we reduce it to a simpler form. After that, FTC can now be applied. The first step to be taken is to make a substitution. For starters, this step will make the integral look less overwhelming. The integrand here is the function

f (x)

4 (1 + 2x)3

=

Using a new variable u (the letter u is more commonly used for substitutions), we let

u = 1 + 2x

[STAGE 1]

So that the integral becomes

∫

4 dx u3

∫ 4u

OR

–3

dx

[STAGE 2]

Now we're getting somewhere! However, something is out of place in the integral

∫ 4u

–3

dx

At this point, we are integrating with respect to the new variable u, thus the symbol dx has to be replaced. We do this by making a second substitution by making a reference to the first substitution in stage 1:

u = 1 + 2x We differentiate u with respect to x:

du/dx = 2

[STAGE 3]

Next, make dx the subject:

dx = ½ du

[STAGE 4]

Then we go back to stage 2: replace dx with ½ du:

∫ (4u

–3

) × ½du

From the properties of integrals, ½ is a constant. Therefore we rewrite the integral:

½

∫ (4u

–3

) du

[STAGE 5]

At this point, you should see how this method works. In this example, by introducing a new variable u, we have reduced

∫

4 (1 + 2x)3

to

dx

½

∫ (4u

–3

) du

Now we can apply the antidifferentiation formula:

½

∫ (4u

–3

) du =

½ (– 2u –1

=

u

2

–2

) + C

+

C

We then put back the original value of u. Hence, the final answer is:

–1 (1 + 2x)2

+

C

Therefore,

∫

4 (1 + 2x)3

dx

–1 (1 + 2x)2

=

+

C

Done! This is how the substitution rule works. Remember, when faced with a relatively complex integral, reducing it to a much simpler form is one good way of evaluating it. The substitution rule states (and I quote from my textbook): “it is permissible to operate with du and dx after the integral signs as if they were differentials”. From the outline of the steps involved in the substitution rule, stage 1 is probably the hardest. In other words, when using the substitution rule to evaluate an integral, the first step is perhaps the most difficult: making the appropriate substitution. Generally, you can follow these guidelines to make an accurate substitution: Try choosing to u represent the complicated part of the integrand (in the example above, u was chosen to represent the denominator because it's obviously more complex than the numerator). Another way is to make u represent a function whose derivative also occurs in the integral (except for a constant factor). Bear in mind that your first substitution may not work. When that happens, simply try another substitution. You can always confirm your answer by differentiating it (which should result in the original integrand in question). Let's study more examples:

EXAMPLE 1 Evaluate the integral by making the given substitution:

∫ x(4 + x )

2 10

u = 4 + x2

dx

Solution First, we rewrite the integral as

∫ (4 + x )

2 10

We

then

make

i

x dx

the substitution u = 4 + x ,so that the integral becomes

∫ (u)

10

x dx

2

ii

Next, we have to replace x dx,and we do that by differentiating u = 4 + x2 to get du/dx

=

2x. We then rewrite

this to give

x dx

=

½ du

Putting this new substitution back into the integral gives

∫ (u)

½ du

10

½

=

∫ (u)

du

10

Now we can use the antiderivative formula to evaluate:

½

∫

(u)10 du

=

u 11 11

½

u 11

=

22

We put back the original value of u to get

∫

x(4 + x2)10 dx

(4 + x2)11

=

+

22

C

EXAMPLE 2 Evaluate the integral by making the given substitution:

∫ x √(x 2

3

u = x3 + 1

+ 1) dx

Solution We rewrite as

∫ √(x

3

+ 1) x2 dx

and using the provided substitution, we have

∫ √(u) x

differentiating u gives du/dx

2

dx =

3x2. We then rewrite this differential to give x2dx

Putting this back in the integral gives

∫ √(u) (1/3) du

=

∫ √(u) du

1/3

Using the general antiderivative formula, we evaluate the integral to give

∫ √(u) du

=

∫ √(u) du

=

1/3

1/3

1 2 u 3/2 3 3 2 u 3/2 9

+

C

which is equivalent to

∫

x2√(x3 + 1) dx

=

2 3 3/2 9 (x + 1)

+

C

=

(1/3) du.

EXAMPLE 3 Evaluate the integral by making the given substitution:

∫

sin√x

dx

√x

u = √x

Solution Putting u = √x in the integral gives

∫

sin u u

i

dx

Also,

ii

du/dx = 1/(2√x) and from (ii),

2du = dx/√x

OR

2du = dx/u

We put this back in (i)to give

∫ sin u

× 2du

=

2

∫ sin u

du

From the antiderivative formula,

2

∫ sin u

du

2

∫ sin u

du

2(– cos u)

=

=

Which equals

Therefore,

∫

sin√x

dx

√x

= =

– 2 cos √x

– 2 cos √x

+ C

EXAMPLE 4 Evaluate the integral by making the given substitution:

∫ sin θ

cosθ dθ

3

u = sin θ

Solution We can rewrite the indefinite integral as

∫ (sinθ)

3

cosθ dθ

i

So that when we put u = sin θ, we get this:

∫u

3

cosθ dθ

ii

– 2 cos u

Differentiating u = sin θ gives du/dθ = cos θ. We rewrite it so that

iii

dθ = du/cos θ We put (iii) back in (ii) to give

∫u

3

du cos θ

cos θ ×

Crossing out the like terms results in

∫u

du

3

Using the antiderivative formula,

∫

u 3 du

=

u4 4

=

=

¼ u4

¼ (sin θ)4

=

¼ sin

4

θ

Therefore,

∫

sin 3θ cosθ dθ

=

¼ sin

4

θ

+

C

EXAMPLE 5 Evaluate the integral

1 + 4x

∫ √1 + x + 2x

dx

2

Solution The first task is to identify the seemingly complex part of the integral. In this case, the denominator is the complex part. Thus, we apply the substitution rule to the denominator. To that effect, we start by letting

u = 1 + x + 2x2 So that the integral becomes

∫

1 + 4x dx √u

i

We differentiate to u = 1 + x + 2x2 to give

du/dx

ii

= 1 + 4x

You'll notice that this derivative just happens to appear in the integral (the numerator to be exact). So, from (ii),

dx

=

du/(1 + 4x)

We put this back in (i) to give

∫

1 + 4x √u

Crossing out the like terms gives

×

du 1 + 4x

∫

du

1

∫ √u

=

√u

=

du

∫u

–½

du

We then use the antiderivative formula which gives

∫

u

–½

du

=

2√u + C

Which equals

1 + 4x

∫ √1 + x + 2x

dx

2

=

2√1 + x + 2x2 + C

Here are two useful hints you can apply when using the substitution rule to evaluate an integral: If an integral contains a square-root, you can start by choosing u to represent the expression under the root. If an integral contains a bracketed expression (integrals that are relatively complex usually do, generally speaking), start by choosing u to represent the expression in brackets.

EXAMPLE 6 Evaluate the integral

∫

(1 – 2y)1.3 dy

Solution We let u = 1 – 2y, so that the integral becomes

∫

u

1.3

i

dy

Differentiating u = 1 – 2y gives

du/dx

=

–2

and so

dy

=

–½ du

ii

× –½ du

iii

Putting (ii) back in (i) gives

∫u

1.3

Using the properties of integrals, we rewrite (iii) above as

–½

∫u

1.3

dy

We can then apply the antidifferentiation formula:

–½

∫u

1.3

=

dy

u

–½

2.3

+ C

4.6

=

+ C

1.3 + 1

–u

=

1.3 + 1

– (1 – 2y)

=

4.6

Therefore,

∫

(1 – 2y)1.3 dy

– (1 – 2y)

=

–½

u

2.3

2.3

+ C

2.3

+ C

2.3

+ C

4.6

EXAMPLE 7

Evaluate the integral

∫ t sin (t ) dt 2

Solution We let u = t2, so that the integral becomes

∫

i

t sin u dt

Observe that t is in a slightly awkward position, we therefore rewrite it as

∫ sin u × t dt Differentiating u = t2 equals du/dt

∫ sin u × ½ du

=

ii 2t and so t dt = ½du. We put this back in (ii) which results in

=

½

∫ sin u du

Then we use the antiderivative formula which gives

∫ sin u du

=

½ (– cos u) + C

∫ t sin (t ) dt

=

–½ cos (t 2) + C

½

=

–½ cos (t 2) + C

Therefore, 2

Evaluate the integral

EXAMPLE 8

∫ sec x tan x √1 + sec x dx Solution We can split the integrand into two parts:

AND

√1 + sec x

sec x tan x

It's easy to find the integral of sec x tan x , but the same cannot be said for .

√1 + sec x Therefore, we let

u

√1 + sec x

i

sec x tan x √u dx

ii

=

which results in

∫

Differentiating u(x) gives du/dx

dx

=

sec x tan x so that

du sec x tan x

=

iii

We put (iii) back in (ii) to give

∫

du sec x tan x

sec x tan x √u ×

Crossing out the like terms leaves

∫ √u du

=

∫u

½

du

which can be easily be evaluated!! Using the antiderivative formula,

∫

u

½

du

=

u

½ + 1

½+1

+ C

=

u 3/2 3/2

+ C

=

2 u 3

3/2

+ C

Therefore,

∫ sec x tan x √1 + sec x dx

=

+ C

EXAMPLE 9

Evaluate the integral

∫ cos

2 (1 + sec x) 3/2 3

4

x sin x dx

Solution This might seem a bit tough, but once you understand that cos easy. So, we rewrite th integral as

∫ (cos

x)4 sin x dx

and let u = cos x, so that the integral becomes

4

x = (cos x)4, evaluating this integral should be

∫u

4

Differentiating u(x) gives du/dx

dx

i

sin x dx =

– sin x so that

du – sin x

=

ii

Substituting this back in (i) results in

∫

du – sin x

u 4 sin x ×

=

–

∫u

du

4

Using the antiderivative formula,

–

∫

u 4 du

= –

=

–u5 + C = 5 1 (cos x)5 + C 5

–

1 5

=

u5

–

+ C 1 (cos 5x) 5

+ C

So,

∫ cos

4

=

x sin x dx

–

1 (cos 5x) 5

+ C

EXAMPLE 10 Evaluate the integral

∫ √ax + 2bx + c ax + b

dx

2

Solution Understand that we are integrating with respect to x, which should tell you that the numbers a and b are constants. The denominator is obviously the complicated part of the integrand, so we let

i

u = ax2 + 2bx + c So that we have

∫

ax + b dx √u

ii

Differentiating (i) gives

du dx

=

2ax + 2b

=

2(ax + b)

which means

dx

=

du 2(ax + b)

Substituting (iii) into (ii) gives

iii

∫

½

ax + b

×

√u

du 2(ax + b)

Crossing out the like terms leaves

∫

1 u½

× ½ du

∫

=

½ u

-½

du

Using the antiderivative formula,

½

∫u

-½

=

du

× 2√u)

(½

=

+

=

C

√ax2 + 2bx + c

+

√u

+

C

+

C

C

Thus,

∫ √ax + 2bx + c ax + b

dx

=

2

√ax2 + 2bx + c

EXAMPLE 11

Evaluate the integral

∫ sec

3

x tan x dx

Solution At this point we understand that the idea behind the substitution rule is to reduce a relatively complex integral to a simple form that can be evaluated. The aim of this specific example is to show that sometimes, it helps to simplify the integrand itself to make the process easier. In the integral above, the integrand is

f(x) =

sec

3

x tan x

which, when factorized, equals

=

f(x)

(sec x)3 tan x

=

(sec x)2 sec x tan x

Clearly, it will be a whole lot easier to evaluate the integral in this form:

∫ (sec x)

2

sec x tan x dx

i

We let u = sec x, which gives

∫u

2

sec x tan x dx

ii

So that

du dx

=

sec x tan x

Putting (iii) into (ii) gives

which gives

dx

=

du sec x tan x

iii

∫

du sec x tan x

u 2 sec x tan x ×

which reduces everything to

∫u

2

du

Using the antiderivative formula,

∫u

2

1 (sec x)3 3

1 u3 + C = 3 1 (sec3 x) + C 3

=

du

=

+ C

Therefore,

∫ sec

3

x tan x dx

1 (sec3 x) 3

=

+ C

EXAMPLE 12 Evaluate the integral

∫

3

√x3 + 1 x5 dx

Solution

This example will demonstrate another useful hint: factorizing a part of the integrand. Of course, it's similar to example 11, except that there will be two substitutions instead of the conventional single substitution. We find that the integrand is

3

√ x3 + 1 x5

f(x) =

which is a product of two functions. For obvious reasons, we should let u = x3 + 1. Now, we should factorize x5 in such a way that one of the factors will also occur in the derivative of u. In this case, the best option is to split x5 into

x3 x2. Thus, we rewrite the integral as

∫

3

√ x3 + 1 x3 × x2 dx

=

∫

3

√ u x3 × x2 dx

ii

Don't forget that u = x3 + 1 which means

du dx

=

3x2

du 3

thus

=

and so,

x2dx

=

du 3

iii

From (ii),

x3

= u –1

iv

Therefore, we substitute (iii) and (iv) into (i) to give

x2dx

i

∫ √ u (u – 1) × 3

du

∫u

=

3

(u – 1)

1/3

1

×

3

du

which results in

∫u

1 3

(u – 1) du

1/3

1 3

=

∫ (u

4/3

– u 1/3) du

At this point, we can apply the antiderivative formula:

1 3

∫

(u 4/3 – u 1/3) du

=

u

1 3

4/3 + 1

u

–

(4/3) + 1

1/3 + 1

+ C

(1/3) + 1

which equals

1 3

∫

1 3

=

3u 7/3 7

–

1 3

=

(u 4/3 – u 1/3) du

3u 4/3 4

u 7/3 7/3

–

u 7/3 4/3 u 7/3 7

=

+ C

+ C

–

u 4/3 4

+ C

Therefore,

∫

3

=

√x3 + 1 x5 dx

(x3 + 1)7/3 7

–

(x3 + 1)4/3 + C 4

EXAMPLE 13 Evaluate the integral

∫x

a

√ b + cxa + 1

dx

c ≠ 0 , a ≠ –1

Solution First, we rewrite the integral:

∫ √ b + cx

a+1

i

xa dx

Keep in mind that a, b, and c are numbers (i.e., constants). We let

u = b + cxa + 1 So that we have

∫ √u x

a

ii

dx

We find that

du dx

=

0 + (a + 1)cx

( a + 1) – 1

=

(a + 1)cxa

iii

Therefore

du

=

xa dx

1

c(a + 1)

=

du

c(a + 1)

iv

Putting (iv) back in (ii) gives

∫ √u

1 du c(a + 1)

×

1

=

c(a + 1)

∫ √u

du

Using the antiderivative formula,

∫ √u

1 c(a + 1)

=

du

2u 3/2 3

1 c(a + 1)

=

1 c(a + 1)

+ C

u 3/2 3/2

+ C

2u 3/2 3c(a + 1)

=

+ C

Therefore,

∫x

a

√ b + cxa + 1

dx

2(b + cxa + 1 )3/2 3c(a + 1)

=

+ C

as long as c ≠ 0 , a ≠ –1

EXAMPLE 14 Evaluate the integral

∫

x (x + 2) ¼

dx

Solution By putting u = x + 2, we have

∫

Then,

du dx

=

x u¼

1

dx

so

i du = dx

Therefore (i) becomes

∫

x u¼

du

ii

We need to do something about the numerator of the integrand above before we can move further. Another substitution needs to be made, and we make use of the first substitution equation: Since u = x + 2, therefore x = u – 2, Thus, (ii) becomes

∫

u–2 u

¼

du

iii

You can see that we've made use of not just one, but two substitutions to arrive at (iii). Now we use basic algebra to simplify the integral before using the antiderivative formula:

∫

u–2 du u¼

∫ (u – 2)(u

=

–¼

=

) du

∫ (u

¾

– 2u

–¼

) du

At this point, I believe we're good to go. Using the antiderivative formula, we have

∫

(u

¾

– 2u

u 7/4 7/4

=

4u

=

7/4

7

–¼

u¾+1 (¾) + 1

=

) du

–

¾ 2 u ¾

–

8u ¾ 3

–

2

4u

=

+ C

u– ¼ + 1 (– ¼) + 1

7/4

7

–

+ C

2

4u ¾ 3

+ C

+ C

Therefore,

∫

x (x + 2) ¼

dx

4(x + 2)

=

7/4

7

–

8(x + 2) 3

¾

+ C

EXAMPLE 15 Evaluate the integral

∫ √1 – x x2

dx

Solution We put u = 1 – x , so that we have this

∫

x2 dx √u

i

Since u = 1 – x , then x = 1 – u and so x2 = (1 – u)2. Therefore (i) becomes

∫

(1 – u)2 √u

dx

=

∫

(1 – u)2 u½

dx

Since u = 1 – x , then

du dx

= –1

Thus, (ii) becomes

so that

– du = dx

=

∫

1 – 2u + u 2 u½

dx

ii

–

∫

1 – 2u + u 2 u½

du

–

=

∫ (1 – 2u + u )u 2

du

–½

Using basic algebra to simplify the integral, we get

–

∫ (1 – 2u + u )u 2

du

–½

–∫

=

(u

– 2u

–½

+ u

½

3/2

) du

Using the antiderivative formula,

–∫ =

(u

– 2u

–½

u½ ½

–

=

½

– 2u

+ u

4u 3/2 3

+

u –½ + 1 (–½) + 1

–

=

) du

2u 3/2 3/2

–

½

3/2

+

u 5/2 5/2

–

2u 5/2 + C 5

+ C

=

2u ½ + 1 (½ ) + 1

–

–

2u

4u 3/2

=

3

½

–

–

u 3/2 + 1 (3/2) + 1

+

4u 3/2 3

2u 5/2 5

+ –

2u 5/2 5

2u

½

+ C

+ C

+ C

Therefore,

∫ √1 – x x2

dx

4(1 – x)3/2 3

=

4 (1 – x)3/2 3

=

2(1 – x)5/2 5

– –

2 (1 – x)5/2 5

– –

2(1 – x)

2(1 – x)

+ C

½

+ C

½

EXAMPLE 16

Evaluate the indefinite integral. Illustrate and check that your answer is reasonable by graphing both the function and its antiderivative (take c=0).

∫

3x – 1 dx (3x2 – 2x + 1)4

Solution 2

We put u = 3x – 2x + 1. Then the integral becomes

∫

3x – 1 u4

dx

i

Next, we compute du/dx:

du dx

=

6x – 2

=

2(3x – 1)

so that

dx

=

du 2(3x – 1)

Putting (ii) back in (i) gives

∫

3x – 1 u4

×

du 2(3x – 1)

=

∫ 2u1

4

du

=

½

∫u

–4

du

ii

Using the antiderivative formula,

½

∫

u

du

–4

=

u –4 + 1 –4+1

½

+ C

=

½

u –3 –3

+ C

This results in

u –3 –6 + C

=

1 – 6u 3

+ C

dx

=

Therefore,

∫

3x – 1 (3x – 2x + 1)4 2

1 – 6(3x2 – 2x + 1)3

+ C

If we graph the integrand of the indefinite integral we just evaluated above, we get this:

Understand that F' = f .It is this concept that we'll apply throughout in sketching the antiderivative F of f. ➢

From the graph above, observe that f(x) → −∞ as x → 0. Therefore, the graph of F will follow a similar pattern.

➢

f(x) is negative on the interval -∞ < x < 1/3. Therefore F will be decreasing on this interval. Also, we find that the y-intercept of F is –1/6.

➢

Observe that f(x) = 0 when x = 1/3. Thus F will have a horizontal tangent at x = 1/3.

➢

f(x) is positive for x > 1/3. Therefore F is increasing from this point onwards.

➢

Notice that f(x) → 0 as x → ∞. This means that F becomes flatter as x → ∞.

When we combine all this information, we obtain something like this: Here's a graph of f and F:

EXAMPLE 17 Evaluate the indefinite integral. Illustrate and check that your answer is reasonable by graphing both the function and its antiderivative (take c=0).

âˆŤ sin

3

x cos x dx

Solution We rewrite the integral as

∫ (sin x)

3

cos x dx

and let u = sin x. Then the integral becomes

∫u

3

cos x dx

i

Next,

du dx

=

cos x

which means

du = cos x dx

Thus, (i) becomes

∫u

3

du

=

¼ u

4

+ C

Therefore,

∫ sin

3

x cos x dx

=

¼ (sin

4

x)

+ C

Recall that

f(x) = sin 3x cos x dx

and

F(x) = ¼ (sin

4

x)

+ C

In this case, we're dealing with f on a particular interval [0, π]. On this interval, observe that f is positive on [0, π/2]. F will therefore increase on this interval. f(x) = 0 when x = π/2. Thus, F will have a horizontal tangent as this point. On the interval [π/2, π], f is negative. This means F will be decreasing on this interval. Finally, observe that

f(x) → 0 as x → π. Therefore, F will be flatter as it approaches the x-axis. Using the information above, we have the graphs of f and F above.

EXAMPLE 18

Evaluate the indefinite integral. Illustrate and check that your answer is reasonable by graphing both the function and its antiderivative (take c=0).

∫ tan θ sec 2

2

θ dθ

Solution

How do we choose what u will represent? Well, we apply the first hint: choosing u to represent a function whose derivative occurs in the integral. With that in mind, we'll choose u = tan θ. Thus, we'll have to rewrite the integral:

∫ (tan θ)

2

sec2 θ dθ

∫

=

u 2 sec2 θ dθ

i

Then,

du dθ

=

sec2 θ

and so

du = sec2 θ dθ

ii

We put (ii) in (i) to give

∫u

2

=

du

1 3 u 3

+ C

Therefore,

∫ tan θ sec 2

2

θ dθ

=

1 (tan 3 θ) + C 3

When you want to graph an antiderivative F using the original function f, always guide yourself with one simple fact: the slope of F equals f (in other words, F'(x) = f(x)). This means that, on ANY interval I, ●

if F'(x) < 0, then f is decreasing on that interval.

●

if F'(x) > 0, then f is increasing on that interval.

Here, we graph the function f(x) = tan2 θ sec2 θ on the interval [-π/5, 6π/5] and discover that

##

●

since f is decreasing on [-π/5,0] and [4π/5, π], it means F will be negative on [-π/5, 0] and [4π/5,π] respectively.

●

The points [0,0] and [π, 0] are the points of inflection## on f. Thus F will follow this pattern.

●

since f is increasing on [0, π/5] and [π, 6π/5], it means F will be positive on [0, π/5] and [π, 6π/5] respectively.

On any curve, a point of inflection is a point where concavity changes, that is, the point where a curve changes

from concave upward or concave downward. Using the information above, we have f and F graphed below:

EXAMPLE 19 Evaluate the indefinite integral. Illustrate and check that your answer is reasonable by graphing both the function and its antiderivative (take c=0).

∫

x dx √x2 + 1

Solution

We let u = x + 1, so that we have 2

∫

x u½

dx

i

Now,

du dx

=

2x

so that

x dx

du 2

=

ii

Therefore (i) becomes

∫u

1 ½

×

du 2

using the antiderivative formula,

=

∫ 2u1

½

du

=

1 2

∫u

–½

du

1 2

∫u

–½

du

=

½ (2u ½) + C

=

u

½

+ C

So,

∫ ●

x dx 2 √x + 1

=

√x2 + 1 + C

Notice that f(x) → 1 as x → ∞, and f(x) → -1 as x → -∞. This implies f that has horizontal asymptotes y = 1 and y = -1 (indicated by the dash lines).

●

Observe that f(x) < 0 on [-∞,0]. This would mean that F is decreasing on this interval.

●

The y-intercept of F is 1.

●

Notice that f(x) > 0 on [0,∞], which implies that F is increasing on this interval.

●

Since F(0) = 0, F will have a horizontal tangent when x = 0 (and this horizontal tangent just happens to be the horizontal asymptote y = 1).

Using the information above, we have this:

So far, we have studied the substitution rule, and seen how it's used to evaluate indefinite integrals. In part three of this tutorial, we examine how this rule is used for definite integrals, and consider some real life applications.

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