evaluatingintegrals(substitution rulei)

EVALUATING INTEGRALS - SUBStitutiONEVALUATING 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:

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

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

[STAGE 2]

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

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:

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

[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

dx ∫

dx 4 u3∫ 4u –3 dx∫OR

(4u –3) du∫ ½

f (x) = 4 (1 + 2x)3

4u –3 dx∫

(4u –3) × ½du∫

Now we can apply the antidifferentiation formula:

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

Therefore,

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 1Evaluate the integral by making the given substitution:

SolutionFirst, we rewrite the integral as

We then make the substitution u = 4 + x2 ,so that the integral becomes

to

½ (– 2u –2) + C=

– 1

u2+ C

– 1 (1 + 2x)2

+ C

=

x(4 + x2)10 dx∫ u = 4 + x2

(4 + x2)10 x dx∫ i

(u)10 x dx∫ ii

4 (1 + 2x)3

dx ∫ (4u –3) du∫ ½

(4u –3) du∫ ½

=

4 (1 + 2x)3

dx ∫ – 1 (1 + 2x)2

+ C

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

Now we can use the antiderivative formula to evaluate:

We put back the original value of u to get


SolutionWe rewrite as

and using the provided substitution, we have

differentiating u gives du/dx = 3x2. We then rewrite this differential to give x2dx = (1/3) du.

Putting this back in the integral gives

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

which is equivalent to

(u)10 ½ du∫ = ½ (u)10 du∫

= u11

11 ½ ½ (u)10 du∫ =

u11

22

x(4 + x2)10 dx∫ = (4 + x2)11

22+ C

x2√(x3 + 1) dx∫ u = x3 + 1

√(x3 + 1) x2 dx∫

√(u) x2 dx∫

√(u) (1/3) du∫ 1/3 √(u) du∫=

1 3

u3/2 2 3

2 9

u3/2 + C

x2√(x3 + 1) dx∫ 2 9 (x3 + 1)3/2 + C

1/3 √(u) du∫ =

1/3 √(u) du∫ =

=


SolutionPutting u = √x in the integral gives

Also,

du/dx = 1/(2√x)

and from (ii),

2du = dx/√x OR 2du = dx/u

We put this back in (i)to give

From the antiderivative formula,

Which equals

Therefore,


SolutionWe can rewrite the indefinite integral as

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

sin√x √x

dx ∫ u = √x

sin u u

dx ∫ i

ii

sin u × 2du∫ = sin u du∫2

sin u du∫2 = 2(– cos u) = – 2 cos u

= – 2 cos √x

= – 2 cos √x + C

sin3θ cosθ dθ∫ u = sin θ

(sinθ)3 cosθ dθ∫ i

u3 cosθ dθ∫ ii

sin u du∫2

sin√x √x

dx ∫

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

dθ = du/cos θ

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

Crossing out the like terms results in

Using the antiderivative formula,

Therefore,

EXAMPLE 5Evaluate the integral

SolutionThe 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

So that the integral becomes

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

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

We put this back in (i) to give

Crossing out the like terms gives

iii

u3 cos θ ×∫ du cos θ

u3 du∫

= ¼ u4 u4 4

¼ (sin θ)4 ¼ sin 4 θ

sin3θ cosθ dθ∫ ¼ sin 4 θ + C

1 + 4x

√1 + x + 2x2 dx ∫

u = 1 + x + 2x2

1 + 4x √u dx ∫ i

du/dx = 1 + 4x ii

dx = du/(1 + 4x)

du

1 + 4x

u3 du∫ = = =

=

1 + 4x √u × ∫

We then use the antiderivative formula which gives

Which equals

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.


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

Differentiating u = 1 – 2y gives

and so

Putting (ii) back in (i) gives

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

We can then apply the antidifferentiation formula:

du √u∫ = du 1

√u ∫ = u –½ du∫

u –½ du∫ = 2√u + C

= 2√1 + x + 2x2 + C

(1 – 2y)1.3 dy∫

u 1.3 dy∫ i

du/dx = –2

dy = –½ du ii

u 1.3 × –½ du ∫ iii

–½ u 1.3 dy∫

1 + 4x

√1 + x + 2x2 dx ∫

Therefore,


SolutionWe let u = t2, so that the integral becomes

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

Differentiating u = t2 equals du/dt = 2t and so t dt = ½du. We put this back in (ii) which results in

Then we use the antiderivative formula which gives

Therefore,


SolutionWe can split the integrand into two parts:

= –½ u 1.3 + 1

1.3 + 1 + C = –½

u 2.3

2.3+ C

– u 2.3

4.6+ C = – (1 – 2y) 2.3

4.6+ C =

=

t sin (t2) dt∫

t sin u dt∫ i

sin u × t dt∫ ii

sin u × ½ du ∫ = ½ sin u du ∫

= ½ (– cos u) + C = –½ cos (t2) + C

=

∫ sec x tan x √1 + sec x dx

–½ u 1.3 dy∫

(1 – 2y)1.3 dy∫ – (1 – 2y) 2.3

4.6+ C

½ sin u du ∫

t sin (t2) dt∫ –½ cos (t2) + C

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

Therefore, we let

which results in

Differentiating u(x) gives du/dx = sec x tan x so that

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

Crossing out the like terms leaves

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

Therefore,


SolutionThis might seem a bit tough, but once you understand that cos 4 x = (cos x)4, evaluating this integral should be

easy. So, we rewrite th integral as

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

√1 + sec x sec x tan x AND

u = i

∫ sec x tan x √u dx ii

dusec x tan x

dx = iii

√u du∫ u ½ du∫=

=u ½ + 1

½ + 1 + C = u 3/2

3/2+ C u 3/2 2

3+ C

= 2 3

(1 + sec x) 3/2 + C

∫ cos 4 x sin x dx

∫ (cos x)4 sin x dx

√1 + sec x

√1 + sec x

∫ sec x tan x √u ×du

sec x tan x

u ½ du∫ =

∫ sec x tan x √1 + sec x dx

Differentiating u(x) gives du/dx = – sin x so that

Substituting this back in (i) results in


So,


SolutionUnderstand 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

So that we have

Differentiating (i) gives

which means

Substituting (iii) into (ii) gives

∫ u4 sin x dx i

du

– sin xdx = ii

=∫ u4 sin x × du– sin x – ∫ u4 du

= – u 5 5

+ C = 1

5u5 + C –

= (cos x)5 + C 15

– (cos 5x) + C 15

–=

=

∫ ax + b

√ax2 + 2bx + c dx

u = ax2 + 2bx + c i

∫ ax + b

√u dx ii

dudx

2ax + 2b = 2(ax + b)=

du2(ax + b)

dx = iii

– ∫ u4 du

∫ cos 4 x sin x dx (cos 5x) + C 15

–

Crossing out the like terms leaves


Thus,


SolutionAt 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

which, when factorized, equals

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

We let u = sec x, which gives

So that

Putting (iii) into (ii) gives

1 u½ ∫ × ½ du =

½

(½ × 2√u) + C = = √u + C

= √ax2 + 2bx + c + C

=

∫ sec 3 x tan x dx

f(x) = sec 3 x tan x

f(x) = (sec x)3 tan x = (sec x)2 sec x tan x

(sec x)2 sec x tan x dx ∫ i

u2 sec x tan x dx ∫ ii

dudx

sec x tan x= dusec x tan x

dx =which gives iii

∫ ax + b

√u × du

2(ax + b)

½ u -½ du ∫

∫ ax + b

√ax2 + 2bx + c dx √ax2 + 2bx + c + C

½ u -½ du ∫

which reduces everything to


Therefore,


SolutionThis 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

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

Don't forget that u = x3 + 1

which means

and so,

From (ii),

x3 = u – 1

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

u2 du ∫

= 13

u3 + C =13

(sec x)3 + C

= 13

(sec3 x) + C

u2 sec x tan x × ∫ dusec x tan x

u2 du ∫

∫ sec 3 x tan x dx = 13

(sec3 x) + C

∫ √x3 + 1 x5 dx3

√ x3 + 1 x5 f(x) =3

∫ √ x3 + 1 x3 × x2 dx i=3 ∫ √ u x3 × x2 dx

3

ii

dudx = 3x2 thus du

3 = x2dx

du3

=x2dx iii

iv

which results in

At this point, we can apply the antiderivative formula:

which equals

Therefore,


SolutionFirst, we rewrite the integral:

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

So that we have

We find that

∫ √ u 3

×(u – 1) du

3= ∫ u1/3 ×(u – 1)

1

3du

=∫ u1/3 (u – 1) du13 ∫(u4/3 – u1/3) du

13

= u 4/3 + 1

(4/3) + 1 –

u 1/3 + 1

(1/3) + 1 + C

13∫(u4/3 – u1/3) du

13

= u7/3 7/3

u7/3 4/3 – + C

13

= 3u7/3 7

– + C 13

3u4/3 4

= u7/3 7

u4/3 4

– + C

∫(u4/3 – u1/3) du13

= (x3 + 1)7/3

7 – + C (x3 + 1)4/3

4∫ √x3 + 1 x5 dx

3

∫ xa √ b + cxa + 1 dx c ≠ 0 , a ≠ –1

∫ √ b + cxa + 1 xa dx i

u = b + cxa + 1

∫ √u xa dx ii

dudx = 0 + (a + 1)cx (a + 1) – 1 = (a + 1)cxa iii

Therefore

Putting (iv) back in (ii) gives


Therefore,


SolutionBy putting u = x + 2, we have

Then,

Therefore (i) becomes

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

du

c(a + 1)xa dx =

=1

c(a + 1)du iv

∫ √u × 1 c(a + 1)

du =1

c(a + 1) ∫ √u du

= 1 c(a + 1)

2u3/2 3 + C = 2u3/2

3c(a + 1) + C

= 1 c(a + 1)

u3/2

3/2 + C 1

c(a + 1) ∫ √u du

= 2(b + cxa + 1 )3/2

3c(a + 1) + C as long as c ≠ 0 , a ≠ –1 ∫ xa √ b + cxa + 1 dx

x

(x + 2) ¼ dx ∫

x u¼

dx ∫ i

dudx = 1 du = dxso

ii

u – 2

u ¼ du ∫ iii

x u¼

du ∫

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:

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

Therefore,


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

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

Since u = 1 – x , then

Thus, (ii) becomes

= 2u 7/4 7/4

– u ¾ ¾

+ C = 24u 7/4 7

– 4u ¾ 3

+ C

= 4u 7/4 7

– 8u ¾

3 + C

∫ (u – 2)(u –¼) du= = ∫ (u ¾ – 2u –¼) du u – 2 u ¼

du ∫

= 2u ¾ + 1 (¾) + 1

– u– ¼ + 1 (– ¼) + 1 + C ∫ (u ¾ – 2u –¼) du

= 4(x + 2) 7/4

7 –

8(x + 2) ¾

3 + C

x

(x + 2) ¼ dx ∫

x2 dx ∫ √1 – x

i

(1 – u)2 dx ∫ √u = (1 – u)2 dx ∫ u ½

ii

dudx = –1 – du = dxso that

1 – 2u + u2 dx ∫ u ½ =

x2 dx ∫ √u

Using basic algebra to simplify the integral, we get


Therefore,

EXAMPLE 16Evaluate the indefinite integral. Illustrate and check that your answer is reasonable by graphing both the function

and its antiderivative (take c=0).

SolutionWe put u = 3x2 – 2x + 1. Then the integral becomes

Next, we compute du/dx:

Putting (ii) back in (i) gives

1 – 2u + u2 du ∫ u ½ – = ∫ (1 – 2u + u2)u – ½ du–

∫ (1 – 2u + u2)u – ½ du– = ∫ (u – ½ – 2u ½ + u 3/2) du–

= ∫ (u – ½ – 2u ½ + u 3/2) du– – u –½ + 1 (–½) + 1

– 2u ½ + 1 (½ ) + 1 + C + u 3/2 + 1

(3/2) + 1

= – u ½

½ – 2u3/2

3/2 + C + u5/2

5/2 – – 4u3/2 3

+ C +2u ½ 2u5/2 5

= – 4u3/2

3+ C – 2u ½ 2u5/2

5+ =

4u3/2

3 – 2u5/2

5– 2u ½ + C

= 4(1 – x)3/2

3 – 2(1 – x)5/2

5– 2(1 – x) ½ + C

= (1 – x)3/243 – 2

5(1 – x)5/2 – 2(1 – x) ½ + C

=

x2 dx ∫ √1 – x

dx ∫ 3x – 1 (3x2 – 2x + 1)4

dx ∫ 3x – 1 u4 i

du

dx = 6x – 2 = 2(3x – 1)du

2(3x – 1)dx =so that ii

× ∫ 3x – 1 u4

du2(3x – 1) = ∫ 1

2u4 du = ∫ u – 4 du½


This results in

Therefore,

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:

∫ u – 4 du½ = u–4 + 1 – 4 + 1 + C = u–3

– 3+ C

= u–3 – 6 + C

1 – 6u3 + C

dx ∫ 3x – 1 (3x2 – 2x + 1)4 = 1

– 6(3x2 – 2x + 1)3 + C

½ ½



∫ sin 3x cos x dx

SolutionWe rewrite the integral as

and let u = sin x. Then the integral becomes

Next,

Thus, (i) becomes

Therefore,

Recall that

∫ (sin x)3 cos x dx

∫ u3 cos x dx

dudx = cos x which means du = cos x dx

i

∫ u3 du = ¼ u 4 + C

∫ sin 3x cos x dx = ¼ (sin 4 x) + C

f(x) = sin 3x cos x dx F(x) = ¼ (sin 4 x) + Cand

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.



SolutionHow 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:

Then,

We put (ii) in (i) to give

Therefore,

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:

∫ tan2 θ sec2 θ dθ

∫ (tan θ)2 sec2 θ dθ ∫ u2 sec2 θ dθ= i

dudθ = sec2 θ and so du = sec2 θ dθ ii

∫ u2 du = u3 + C 13

= (tan3 θ) + C 13∫ tan2 θ sec2 θ dθ



SolutionWe let u = x2 + 1, so that we have

Now,

Therefore (i) becomes

using the antiderivative formula,

∫ x dx √x2 + 1

x dx ∫ u ½ i

dudx = 2x so that x dx =

du2

ii

= ×∫ 1u ½

du2

= du∫ 12u ½

12 ∫ u – ½ du

So,

● 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.

calculus4engineeringstudents.com

= √x2 + 1 + C

= ½ (2u ½) + C = u ½ + C12 ∫ u – ½ du

∫ x dx √x2 + 1

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