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Integration by Substitution: The Fundamental Theorem of Calculus demonstrated the importance of being able to find anti-derivatives. We now introduce some methods for finding anti- derivatives: 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 This method is sometimes also referred to as u-substitution. This is really the chain-rule for differentiation being applied in reverse order. Example: Evaluate Z e 4x dx We note that the derivative of e 4x is e 4x , but differentiating e 4x will also give rise to an additional factor of 4 because of the chain rule, so to see this problem better we make this substitution: u =4x From this substitution we then have: du dx =4 The substitution method says that we can treat du, dx as differentials, in other words, we can write the above as: du =4 dx You can look at the substitution method as a way of recognizing the antiderivative of the result of the chain rule. Treating du, dx as differentials allow us to rewrite every term inside the integral sign, including dx, to be in terms of u and du. u =4x du =4 dx dx = 1 4 du Z e 4x dx = Z e u 1 4 du = 1 4 Z e u du = 1 4 e u + C Since the original question is in terms of x, our answer must also be in terms of x, so we back substitute to get: Z e 4x dx = 1 4 e u + C = 1 4 e 4x + C Notice that with the substitution method (or simply by inspection), we see that if the anti-derivative of a function f (x) can be found, then the anti-derivative of a function of the form f (cx), where c is any constant, can also be found. In fact, if g (x) is the anti-derivative of f (x), then the anti-derivative of f (cx) will be 1 c · g (cx) For example, since Z e x dx = e x + C , so Z e 5x dx = 1 5 e 5x + C

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Page 1: The Fundamental Theorem of Calculus demonstrated …wjeh/math110B/int_method/calculus_note... · The Fundamental Theorem of Calculus demonstrated ... as u-substitution. This is really

Integration by Substitution:

The Fundamental Theorem of Calculus demonstrated the importance of beingable to find anti-derivatives. We now introduce some methods for finding anti-derivatives:

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

This method is sometimes also referred to as u-substitution.

This is really the chain-rule for differentiation being applied in reverse order.

Example: Evaluate

∫e4x dx

We note that the derivative of e4x is e4x, but differentiating e4x will also giverise to an additional factor of 4 because of the chain rule, so to see this problembetter we make this substitution:

u = 4x

From this substitution we then have:du

dx= 4

The substitution method says that we can treat du, dx as differentials, in otherwords, we can write the above as: du = 4 dx

You can look at the substitution method as a way of recognizing the antiderivativeof the result of the chain rule.

Treating du, dx as differentials allow us to rewrite every term inside the integralsign, including dx, to be in terms of u and du.

u = 4x⇒ du = 4 dx⇒ dx =1

4du∫

e4x dx =

∫eu(

1

4

)du =

1

4

∫eu du =

1

4eu + C

Since the original question is in terms of x, our answer must also be in terms ofx, so we back substitute to get:∫e4x dx =

1

4eu + C =

1

4e4x + C

Notice that with the substitution method (or simply by inspection), we see thatif the anti-derivative of a function f(x) can be found, then the anti-derivativeof a function of the form f(cx), where c is any constant, can also be found. Infact, if g(x) is the anti-derivative of f(x), then the anti-derivative of f(cx) will

be1

c· g(cx)

For example, since

∫ex dx = ex + C, so

∫e5x dx =

1

5e5x + C

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Similarly,

∫sinx dx = − cosx+ C, so

∫sin(8x) dx = −1

8cos(8x) + C

Keep in mind that this works only if c is a constant.

It is important to note that when you use substitution, you must account foreverything in the original variable. If you substitute u in terms of x, youmust also account for dx in terms of u and du. Most of the time, dx does notjust change into du, but instead, you will need to differentiate and treat them asdifferentials/numbers.

Example:

Find

∫2x sin(x2) dx

Let u = x2, then du = 2x dx

and the above integral becomes:∫2x sin(x2) dx =

∫sin(u) du = − cosu+ c = − cos(x2) + c

Example:

Find

∫x

x2 − 1dx

Let u = x2 − 1, then du = 2x dx, so x dx =1

2du∫

x

x2 − 1dx =

∫1

2udu =

1

2

∫1

udu =

1

2ln |u|+ c =

1

2ln |x2 − 1|+ c

Example:

Find

∫tan(x) dx

Ans:

∫tan(x) dx =

∫sin(x)

cos(x)dx

Let u = cos(x), then du = − sin(x)dx, so −du = sin(x)dx, and we have∫sin(x)

cos(x)dx =

∫−1

udu = −

∫1

udu = − ln |u|+ c = − ln | cos(x)|+ c

Using the property of log, we know that

r ln(x) = ln(xr), the above expression can be written as:

− ln | cosx|+ C = ln | cosx|−1 + C = ln

(1

| cosx|

)+ C = ln | secx|+ C

Notice in this example that, when use use the substitution u = cos(x), dx does

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not just become du. Instead, dx = − 1

sinxdu. We have the expression sinx dx

inside the integral and that is the reason we change everything in the expressionto be in terms of u and du.

Example:

∫x2 + 1

x+ 2dx

We make the substition u = x+2, then x = u−2⇒ x2 = (u− 2)2, also, du = dx,so the above expression becomes:∫x2 + 1

x+ 2dx =

∫(u− 2)2 + 1

udu =

∫u2 − 4u+ 4 + 1

udu =

∫u2 − 4u+ 5

udu

We can now split the fraction into three fractions and simplify:∫u2 − 4u+ 5

udu =

∫u2

u− 4u

u+

5

udu

=

∫u− 4 + 5

(1

u

)du =

u2

2− 4u+ 5 ln |u|+ C

Rewrite the expression back in terms of x, we get:∫x2 + 1

x+ 2dx =

u2

2− 4u+ 5 ln |u|+ C =

(x+ 2)2

2− 4(x+ 2) + 5 ln |x+ 2|+ C

When you use substition for a definite integral, you have two options. You caneither finish the integration and back substitute the values of x using the limitsof integration, or you can change the limits of integration at the beginning.

Example:∫ e

1

lnx

xdx

Approach 1: Let u = lnx, then du =1

xdx, so∫ e

1

lnx

xdx =

∫ x=e

x=1

u du =u2

2

]x=e

x=1

=(lnx)2

2

]e1

=(ln e)2

2− (ln 1)2

2=

1

2

Notice that in this approach, we do not change the limits of integration when wesubstitute u for x, so we explicitly remind ouself that the limits of integrationare for x, not for u, and after we finish the integration, we substitute back in x.

Approach 2: Let u = lnx, then du =1

xdx. Also, since u = lnx, when x = 1,

u = ln 1 = 0, and when x = e, u = ln e = 1, we change the limits of integrationin terms of u:

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∫ e

1

lnx

xdx =

∫ 1

0

u du =u2

2

]10

=1

2

In this approach the limits of integration is immediately changed together withthe change from x to u, and we do not need to change u back to x once we finishthe problem. Either approach is perfectly valid and you may use whichever youlike.

Integration by Parts:

Suppose f , g are differentiable functions of x, the product rule for differentiationgives:

(f(x) · g(x))′ = f ′(x) · g(x) + g′(x) · f(x)

Taking antiderivative of both sides and using the property of integrals, we have:

f(x) · g(x) =

∫f ′(x) · g(x) dx+

∫g′(x) · f(x) dx

Rearrange the terms we get:∫g′(x) · f(x) dx = f(x) · g(x)−

∫g(x) · f ′(x) dx

Using u to represent f(x) and v to represent g(x), the above formula can bewritten in this more compact form:∫u dv = uv −

∫v du

The above formula is the method of integration by parts. You can think of thismethod as the reverse of the product rule for differentiation. What this methodallows us to do is to change the format of an integral to another one which we(hopefully) can more easily find its derivative.

When applying the method of integration by parts, the original integral starts as∫u dv, where u is a function of x, and dv is another function of x, including the

differential dx. You then turn u into du, and find the function v as a function ofx, given that you know what is dv. The integration by parts formula then says

you have to evaluate a different integral,

∫v du.

Example:

Find

∫xex dx

If we let u = x and dv = ex dx, then

du = dx and v = ex, and the above integral becomes:

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∫xex dx =

∫u dv, using the method of integration by parts, we get:∫

xex dx =

∫u dv

= uv −∫v du

= xex −∫ex dx

= xex − ex + C

Example:

Find

∫lnx dx

We want to use integration by parts to solve this integral. While the integralappears to have only one product, ln x, don’t forget that for any expression,

dv = 1 · dv. We will let u = ln x and let dv = 1 dx, then du =1

xdx, and v = x,

and using integration by parts we have:∫lnx dx =

∫u dv

= uv −∫v du

= xlnx−∫x · 1

xdx

= x lnx−∫

1 dx

= x lnx− x+ C

Example: Evaluate

∫x5√x3 + 1 dx

Let’s try to solve this problem by the method of integration by parts.

First we will decide which factor inside the integrand we will use to be u andwhich will be dv.

If we let u = x5 and dv =√x3 + 1 dx, we will have a hard time figuring out

what is v, since to find dv we will need to first find the anti-derivative of√x3 + 1,

which is not easy to do.

We may try to let u =√x3 + 1 and dv = x5 dx. This way at least we can find v

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using the power rule, and this gives us:

du =3x2

2√x3 + 1

dx v =x6

6,

Using the formula for by parts we get:∫x5√x3 + 1 dx =

√x3 + 1 · x

6

6−∫x6

6· 3x2

2√x3 + 1

dx

=x6

6

√x3 + 1−

∫x8

4√x3 + 1

dx

We end up with an integral that is more complicated than what we started with.

The trick for this problem is, instead of using all 5 powers of x as a factor of u,we notice that the derivative of x3 will give rise to a factor of x2 term (with aconstant factor, but that is not important). What we do instead is to redistributethe powers of x by looking at the integral as:∫x5√x3 + 1 dx =

∫x3(x2√x3 + 1

)dx

And we let:

u = x3, dv = x2√x3 + 1 dx

We can integrate x2√x3 + 1 dx by substitution (use z = x3 + 1) to find v, then

we have:

du = 3x2 dx, v =2

9

(x3 + 1

)3/2,

then the integral becomes:∫x5√x3 + 1 dx =

∫x3(x2√x3 + 1

)dx

=

∫u dv

= uv −∫v du

= x3[

2

9

(x3 + 1

)3/2]− ∫ [2

9

(x3 + 1

)3/2]3x2 dx

The integral on the right can now be evaluated by substitution again, and weget:

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∫x5√x3 + 1 dx =

∫x3(x2√x3 + 1

)dx

=

∫u dv

= uv −∫v du

= x3[

2

9

(x3 + 1

)3/2]− ∫ [2

9

(x3 + 1

)3/2]3x2 dx

=2

9x3(x3 + 1

)3/2 − 4

45

(x3 + 1

)5/2+ C

Example: Evaluate

∫ex sinx dx

We will evaluate by parts. In this example it does not matter which one we useas our u and which one is dv.

Let u = ex, dv = sinx dx

then du = ex dx, v = − cosx

from the formula for integration by parts we have:∫exsinx dx =

∫u dv

= uv −∫v du

= ex(− cosx)−∫

(− cosx)ex dx

= −ex cosx+

∫ex cosx dx

The integral on the right seems as complicated as the one we started with, sowhat’s the difference here? To see what happens, we will use the by parts methodagain, still letting u = ex and this time, dv = cosx dx, then:

du = ex dx, v = sinx

Using by parts again we have:∫ex sinx dx = −ex cosx+

∫ex cosx dx

= −ex cosx+

[ex sinx−

∫(sinx)ex dx

]= −ex cosx+ ex sinx−

∫ex sinx dx

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We end up with the same integral we started with. It may appear that we havegone no where, but notice that the integral on the left has a negative sign, sorearranging the equation gives us:∫

ex sinx dx = −ex cosx+ ex sinx−∫ex sinx dx

2

∫ex sinx dx = −ex cosx+ ex sinx∫ex sinx dx =

1

2(−ex cosx+ ex sinx) + C

Notice that we still need to put in the +C at the end.

Integrals involving Trigonometric Functions:

In finding anti-derivatives involving tri-functions, sometimes we need to changethe expression into different forms using trigonometric identities, and after that,if necessary, use substitution or other methods to finish the integral.

Example: Evaluate

∫sin5 x dx

This is an integral that involves an odd power of sin, we use the fact that thederivatives of sinx is cosx and the pythagorean identity: sin2 x + cos2 x = 1 tochange the above into:∫

sin5 x dx =

∫ (sin4 x

)(sinx) dx=

∫ (sin2 x

)2(sinx) dx=

∫ (1− cos2 x

)2(sinx) dx

Using the substitution u = cosx, we get: du = − sinx dx, and the above integralbecomes:∫ (

1− cos2 x)2

(sinx) dx =

∫ (1− u2

)2(−du)

=

∫−(1− 2u2 + u4) du =

∫−1 + 2u2 − u4 du

This is now a polynomial in u which can be easily evaluated using the power rule:∫−1 + 2u2 − u4 du = −u+ 2

(u3

3

)− u5

5+ C

We now just substitute the expression back to in terms of x:

−u+ 2

(u3

3

)− u5

5+ C = − cosx+

2

3cos3 x− cos5 x

5+ C

Example:

∫sin6 x dx

This time we have an even power of sine. The previous approach we used will

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not work, as when we change factors of sin2 x to 1 − cos2 x, and use one factorsinx to account for du, there will be one factor of sinx left, which will not beaccounted for. Instead, we use another identity from trigonometry to reduce thepowers of sine.

Remember the double angle identity for cosine:

cos(2x) = 2 cos2 x− 1 = 1− 2 sin2 x

If we use the first equation and solve for cos2 x we get:

cos(2x) = 2 cos2 x− 1

2 cos2 x = cos(2x) + 1

cos2 x =1

2(cos(2x) + 1)

Using the second equation and solve for sin2 x we get:

cos(2x) = 1− 2 sin2 x

2 sin2 x = 1− cos(2x)

sin2 x =1

2(1− cos(2x))

These two formulas are the power reduction or half angle formula fromtrigonometry. We can use these formula to reduce a power of sine or cosine.

With this, the above gives:∫sin6 x dx =

∫ (sin2 x

)3dx =

∫ (1

2(1− cos 2x)

)3

dx

=1

8

∫1− 3 cos(2x) + 3 cos2(2x)− cos3(2x) dx

=1

8

[∫1 dx− 3

∫cos(2x) dx+ 3

∫cos2(2x) dx−

∫cos3(2x) dx

]=

1

8

[x− 3

2sin(2x) + 3

∫ (1

2(1 + cos(4x))

)dx−

∫cos2(2x) cos(2x) dx

]Notice that the half-angle identity is used again for the third integral.

=1

8

[x− 3

2sin(2x) +

3

2x+

3

8sin(4x)−

∫(1− sin2(2x)) cos(2x) dx

]=

1

8

[x− 3

2sin(2x) +

3

2x+

3

8sin(4x)−

∫cos(2x)− cos(2x) sin2(2x) dx

]We use the substitution method again for the last integral, with u = sin(2x)

=1

8

[x− 3

2sin(2x) +

3

2x+

3

8sin(4x)−

(1

2sin(2x)− 1

6sin3(2x) + c

)]

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=5

16x− 1

4sin(2x) +

3

64sin(4x) +

1

48sin3(2x) + C

Applying similar methods, by using the pythagorean identity or the half-angleidentity, we can integrate any function of the form sinn x, cosn x, where n couldbe an odd or even number. What if we have products of sine and cosine?

Example: Evaluate:

∫sin3 x cos6 x dx

If we have factors of both sine and cosine in our integral, as long as one of thepowers is odd, we can take one of the factors to be the derivative of the otherfunction, and make a u-substitution with u to be the other function.

In the example above, since sine has an odd power, we use one factor of sinx tobe the derivative of cosx (with a negative sign), and change the rest into cosineby using the pythagorean theorem:∫

sin3 x cos6 x dx =

∫(sin2 x)(cos6 x)(sinx) dx =

∫(1− cos2 x)(cos6 x)(sinx) dx

Let u = cosx, then du = − sinx dx ⇒ sinx dx = −du. The above integralbecomes:∫

(1− cos2 x)(cos6 x)(sinx) dx =

∫(1− u2)(u6) (−du) = −

∫(1− u2)u6 du

= −∫u6 − u8 du = −

(1

7u7 − 1

9u9 + c

)= −1

7cos7 x+

1

9cos9 x+ C

If an integral consist of even powers of sine and cosine, we use the power reduction(half-angle) formula to change all the powers to be just a linear power of cosine.Notice that you may end up having to apply the power reduction formula multipletimes.

Example:

∫sin2 x cos2 x dx

We use the power reduction formula to change both sin2 x and cos2 x to be interms of cos(2x):∫

sin2 x cos2 x dx =

∫ [1

2(1− cos(2x))

] [1

2(1 + cos(2x))

]dx

=

∫1

4(1− cos(2x)) (1 + cos(2x)) dx =

1

4

∫1− cos2(2x) dx

We will use the power reduction formula one more time to reduce the power ofcos2(2x) and the above becomes:

1

4

∫1− cos2(2x) dx =

1

4

∫1−

[1

2(1 + cos(4x))

]dx =

1

4

∫1− 1

2− 1

2cos(4x) dx

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=1

4

∫1

2− 1

2cos(4x) dx =

1

4

(1

2x− 1

8sin(4x) + c

)=

1

8x− 1

32sin(4x) + C

We may also use similar technique and other trigonometric identities to integratefunctions that involve other trigonometric functions:

Example:

∫tan3 x secx dx

We know that the derivative of secx is sec x tanx, so if we let u = secx, thendu = secx tanx dx will use the sec x and one factor of tan x. This will leave uswith tan2 x. We can use another form of the pythagorean identity:

tan2 x+ 1 = sec2 x

to change between tan2 x and sec2 x∫tan3 x secx dx =

∫tan2 x (tanx secx) dx

Let u = secx, then du = secx tanx dx,

and tan2 x+ 1 = sec2 x⇒ tan2 x = sec2 x− 1,

the above becomes:∫tan2 x (tanx secx) dx =

∫(u2 − 1) du =

1

3u3 − u+ C =

1

3sec3 x− secx+ C

In general, if our integral consist of powers of tangent and secant, if the powerof tangent is odd, and there is at least one factor of secant, using a substitutionu = secx and change all the other powers of tangent into in terms of secant willallow us to integrate the function.

Example:

∫tan5 x sec4 x dx

This integral consist of an even power of secants. Since the derivative of tanxgives rise to sec2 x, we use the substitution:

u = tanx⇒ du = sec2 x dx

The above becomes:∫tan5 x sec4 x dx =

∫(tan5 x)(sec2 x) sec2 x dx

=

∫(u5)(u2 + 1) du =

∫u7 + u5 du =

1

8u8 +

1

6u6 + C

=1

8tan8 x+

1

6tan6 x+ C

If an integral consists of an even power of secant and one of more powers of tangentis present, use the substitution u = tanx and the appropriate trigonometric

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identities to change everything into in terms of u.

Example:

∫secx dx

This is a more tricky one. Since the expression is absent of any tangent function,and there is only one secant term, none of the methods we used before will work.Instead, we introduce something extra to the integral:∫

secx dx =

∫secx · secx+ tanx

secx+ tanxdx

Notice that the fractionsecx+ tanx

secx+ tanxis just equal to 1. So we did not change the

expression by multiplying this to the function secx. However, upon multiplyingthis we get the integral to a new form:∫

secx·secx+ tanx

secx+ tanxdx=

∫(secx)(secx+ tanx)

secx+ tanxdx=

∫sec2 x+ secx tanx

secx+ tanxdx

Now we make a substitution of

u = secx+ tanx, then

du = (secx tanx+ sec2 x) dx,

and the above integral becomes:∫sec2 x+ secx tanx

secx+ tanxdx =

∫du

u= ln |u|+ C = ln | secx+ tanx|+ C

In other words,

∫secx dx = ln | secx+ tanx|+ C

Example:

∫sec3 x dx

We start with integration by parts. We know that the integral of sec2 x can beeasily found, so we use the substitution

dv = sec2 x dx⇒ v = tanx

u = secx⇒ du = secx tanx dx

With the formula for by parts, we have:∫sec3 x dx =

∫(secx)(sec2 x) dx = (secx)(tanx)−

∫(tanx)(secx tanx) dx

= secx tanx−∫

(secx)(tan2 x) dx = secx tanx−∫

(secx)(sec2 x− 1) dx

= secx tanx−∫ (

sec3 x− secx)dx = secx tanx−

∫sec3 x dx+

∫secx dx

Notice that this is another example where the original integral, after an applica-

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tion of by parts, shows up again on the right, but with a negative sign. We nowsolve the equation for the integral:∫

sec3 x dx = sec x tanx−∫

sec3 x dx+

∫secx dx

2

∫sec3 x dx = sec x tanx+

∫secx dx∫

sec3 x dx =1

2

(secx tanx+

∫secx dx

)From the previous example we know that∫

secx dx = ln | secx+ tanx|+ C

we can now complete the problem:∫sec3 x dx =

1

2

(secx tanx+

∫secx dx

)=

1

2(secx tanx+ ln | secx+ tanx|) + C

Example:

∫tan3 x dx

We use the pythgorean identiy to change two of the powers of tanx into sec2 xthen use substitution:∫

tan3 x dx =

∫tan2 x tanx dx =

∫(sec2 x− 1)(tanx) dx

=

∫sec2 x tanx− tanx dx =

∫sec2 x tanx dx−

∫tanx dx

To solve the first integral, we make a substitution

u = tanx⇒ du = sec2 x dx

The first integral becomes:∫sec2 x tanx dx =

∫u du =

1

2u2 + C =

1

2tan2 x+ C

From the section on substitution we know that∫tanx dx = ln | secx|+ C

Putting these together gives:∫tan3 x dx =

∫sec2 x tanx− tanx dx

=

∫sec2 x tanx dx−

∫tanx dx =

1

2tan2 x− ln | secx|+ C

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In general, anytime we need to integrate an odd power of tan x, we can turn twoof the powers into sec2 x and reduce the integral to one that involves the nextodd power. A step-wise procedure like this will allow us to integrate tann x ofany odd power.

Since cscx and cotx have basically the same relationship as sec x and tanx, anyintegral that involve cscx and cotx can be solved using similar technique.

Trigonometric Substitution:

For expressions that involve the sum or difference of two squares, or the radical ofsum or difference of two squares, for example, expressions of the form a2 + x2 ora2−x2, or

√a2 − x2 ...etc, one can often use a trigonometric function as substitute

for the expression. It is usually easier to find the integral of the trigonometricfunction than the integral of the original expression.

Example:

∫ √1− x2 dx

Using a right triangle where the hypothenuous is 1, and the opposite side to angleθ is x, we can turn the above expression into one in terms of trig functions:

√1− x2

x1

θ

We can then make the substitution according to the picture:

x = sin θ, cos θ =√

1− x2

differentiate we have: dx = cos θ dθ.

So the integral becomes:∫ √1− x2 dx =

∫cos θ cos θ dθ =

∫cos2 θ dθ =

∫1

2(1 + cos 2θ) dθ

=1

2

(θ +

sin 2θ

2

)+ C =

1

2

(θ +

2 sin θ cos θ

2

)+ C

Notice we used the double angle identity for sine in the above.

=1

2

(sin−1 x+ x

√1− x2

)+ C

A trigonometric substitution may be used anytime your integral involves the sumor difference of two squares. In other words, if you have expressions of the formx2 + a2, or x2 − a2, or a2 − x2, where a is a constant.

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Depending on which form you have, you want to draw the right triangle with thesides labeled accordingly:

If the expression involves x2 + a2, you will want x and a to be the two legs of theright triangle.

a

x

√x2 + a2

θ

For this case, if you labeled the opposite side of θ as x, you will have to integratea function that involves tangent and secant. If you labeled the adjacent side asx, you will be dealing with cotangents and cosecants. While either of these twolabels will work, since we have been dealing with tangent and secant more often,it is a good idea that you label the opposite side x

If the expression involves x2−a2, then x should be the hypotenuses, and a shouldbe one of the two legs.

a

√x2 − a2

x

θ

For this case, labeling the adjacent side to θ as a will give you functions in termsof tangent and secant. Labeling the opposite side as a will give you functions interms of cotangent and cosecant.

If the expression involves a2−x2, then a should be the hypotenuses, and x is oneof the two legs.

√a2 − x2

xa

θ

Example:

∫x5√x2 + 2

dx

In this integral, we are looking at the sum of two squares, by letting a =√

2. Wewant to put x and

√2 on the two legs of the right triangle.

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√2

x

√x2 + 2

θ

According to picture,

tan θ =x√2⇒ x =

√2 tan θ ⇒ dx =

√2 sec2 θ dθ

cos θ =

√2√

x2 + 2⇒ 1√

2cos θ =

1√x2 + 2

The above expression becomes:∫x5√x2 + 2

dx =

∫ (x5)( 1√

x2 + 2

)dx =

∫ (√2 tan θ

)5( 1√2

cos θ

)√2 sec2 θ dθ

=

∫(√

2)5 tan5 θ sec θ dθ =(√

2)5 ∫ (

tan4 θ)

(tan θ sec θ) dθ

=(√

2)5 ∫ (

sec2 θ − 1)2

(tan θ sec θ) dθ

=(√

2)5 ∫ (

sec4 θ − 2 sec2 θ + 1)

(tan θ sec θ) dθ

=(√

2)5 [∫ (

sec4 θ)

(tan θ sec θ) dθ −∫ (

2 sec2 θ)

(tan θ sec θ) dθ +

∫(tan θ sec θ) dθ

]=(√

2)5(1

5sec5 θ − 2

3sec3 θ + sec θ

)+ C

Note from the picture that

sec θ =

√x2 + 2√

2=

√x2 + 2

2=

√x2

2+ 1

we have a back substitution:(√2)5(1

5sec5 θ − 2

3sec3 θ + sec θ

)+ C

=(√

2)5 [1

5

(x2

2+ 1

)5/2

− 2

3

(x2

2+ 1

)3/2

+

(x2

2+ 1

)1/2]

+ C

Same comment for trigonometric substitution as with u-substitution. When youchange the expression in x to a trigonometric function in θ, dx does not justturn into dθ. The dx term in the original integral must be accounted for bydifferentiating both sides of the substitution equation and using differentials.

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Integration by Partial Fractions:

Integrating a rational expression, if the numerator has a higher degree than thedenominator, one can always perform long division to reduce it to a form wherewe have a polynomial and a rational expression with the degree of numerator lessthan degree of denominator.

For example, consider:x3 − 2x2 − x+ 1

x2 + 3

In this rational function, the degree of the numerator is greater than the degreeof the denominator. We can perform a long division of polynomials to rewritethe expression:

x− 2

x2 + 3)

x3 − 2x2 − x + 1− x3 − 3x

− 2x2 − 4x + 12x2 + 6

− 4x + 7

We can then write the rational function in the form:

x3 − 2x2 − x+ 1

x2 + 3= (x− 2) +

−4x+ 7

x2 + 3

We will discuss how we can integrate a rational function where the denominatoris a quadratic, and the degree of the numerator is less than the degree of thedenominator.

What if we have a rational expression where the denominator has a degree greaterthan 2? In general, every polynomial of real coefficient can be factored into prod-ucts of linear and quadratic terms. What we can do is to factor the donominatorof a rational function as much as possible, then split the fraction.

Example:

Evaluate

∫1

x2 + 3x− 4dx

According to the method of partial fractions, we can factor the denominator andexpress the rational function as sums of individual fractions with the factors inthe denominator. In other words, we can rewrite the fraction like this:

1

x2 + 3x− 4=

1

(x+ 4)(x− 1)=

A

x+ 4+

B

x− 1

where A and B are constants.

To find the values of A and B, there are two methods. You may use either one:

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First Method: If the two sides of the above equation are to be equal to eachother, then they must be equal for all values of x. You may pick any valid valuesof x and plug into the equation. With two unknowns (A and B), you need topick two values for x. Any value of x that you pick will work, as long as x 6= −4,x 6= 1. We use small and integral values of x to make the resulting equations assimple as possible:

Set x = 0, we have the equation:

−1

4=A

4−B

Set x = 2, we get:

1

6=A

6+B

Solving for A and B in this system of equation gives us:

A = −1

5, B =

1

5

Second Method: Another approach to find the values of A and B is by addingback the two fractions on the right hand side, and then compare coefficients ofthe two functions.

A

x+ 4+

B

x− 1=A(x− 1) +B(x+ 4)

(x+ 4)(x− 1)=Ax− A+Bx+ 4B

(x+ 4)(x− 1)=

(A+B)x+ (4B − A)

x2 + 3x− 4

Comparing the two polynomials we have: (A+B)x+ (4B − A) = 1

Equating the coefficients, we want A+B = 0 and 4B − A = 1

Solving this system gives:

A = −1

5, B =

1

5

We have:∫1

x2 + 3x− 4dx =

∫−1/5

x+ 4+

1/5

x− 1dx = −1

5ln |x+ 4|+ 1

5ln |x− 1|+ c

Example:

Evaluate

∫x2 + 2x− 1

x3 − xdx

After factoring and splitting the denominator we get:

x2 + 2x− 1

x3 − x=

x2 + 2x− 1

x(x− 1)(x+ 1)=A

x+

B

x− 1+

C

x+ 1

First method:

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Setting x = 2, x = −2, and x = 3 (your choice may be different, but the resultwill be the same) gives the system of equation:

1

2A + B +

1

3C =

7

6

−1

2A − 1

3B − C =

1

61

3A +

1

2B +

1

4C =

7

12

Solving the system of equation gives the solution:

A = 1, B = 1, C = −1

Second method:

=A(x− 1)(x+ 1) +B(x)(x+ 1) + C(x)(x− 1)

x(x− 1)(x+ 1)=Ax2 − A+Bx2 +Bx+ Cx2 − Cx

x(x− 1)(x+ 1)

=(A+B + C)x2 + (B − C)x+ (−A)

x3 − x=x2 + 2x− 1

x3 − xEquating coefficients gives:

A + B + C = 1

B − C = 2

−A = −1

Solving the system of equations give:

A = 1, B = 1, C = −1

We have:∫x2 + 2x− 1

x3 − xdx =

∫1

x+

1

x− 1+−1

x+ 1dx

= ln |x|+ ln |x− 1| − ln |x+ 1|+ c

In the case if the denominator has a non-reduceble quadratic factor, we need alinear factor on the numerator.

Example: Evaluate

∫2

x3 + x2 + xdx

Factoring gives:

∫2

x3 + x2 + xdx =

∫2

x(x2 + x+ 1)dx

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Notice that the quadratic x2+x+1 cannot be further factored by real coefficient,we will have:

2

x(x2 + x+ 1)=A

x+

Bx+ C

x2 + x+ 1

Notice that the numerator is a linear term when the denominator is a quadratic.

First Method:

We use x = 1, x = −1, and x = 2. This gives, respectively, the three equations:

A +1

3B +

1

3C =

2

3

−A − B + C = −2

1

2A +

2

7B +

1

7C =

1

7

Solving the system gives:

A = 2, B = −2, C = −2

Second Method:

A

x+

Bx+ C

x2 + x+ 1=A(x2 + x+ 1) + (Bx+ C)(x)

x(x2 + x+ 1)

=Ax2 + Ax+ A+Bx2 + Cx

x3 + x2 + x=

(A+B)x2 + (A+ C)x+ A

x3 + x2 + x=

2

x3 + x2 + x

Equating coefficients give:

A + B = 0

A + C = 0

A = 2

Solving the system gives

A = 2, B = −2, C = −2

We have:∫2

x3 + x2 + xdx =

∫2

x+−2x− 2

x2 + x+ 1dx

= 2 ln |x| −∫

2x+ 1

x2 + x+ 1+

1

x2 + x+ 1dx

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= 2 ln |x| − ln |x2 + x+ 1| −∫

1

x2 + x+ 1dx

To integrate this last fraction, we use the method of completing the square anduse the formula for the derivative/integral of arctangent∫

1

x2 + a2dx =

1

atan−1

(xa

)+ c

The above integral becomes:∫1

x2 + x+ 1dx =

∫1

(x2 + x+ 1/4) + (3/4)dx =

∫1

(x+ 1/2)2 + (√

3/4)2dx

We now use substitution: u = x+ 1/2, the integral becomes:∫1

u2 + (√

3/4)2du =

1√3/4

tan−1

(u√3/4

)+ c

Back substitute for x and simplify gives:∫1

u2 + (√

3/4)2du =

2√3

tan−1(

2x+ 1√3

)+ c

Finally, putting everything together:∫2

x3 + x2 + xdx = 2 ln |x| − ln |x2 + x+ 1| −

∫1

x2 + x+ 1dx

= 2 ln |x| − ln |x2 + x+ 1| − 2√3

tan−1(

2x+ 1√3

)+ c

In the case when the denominator consist of repeated linear factors, each factorand repeated factor needs to be accounted for:

Example: Evaluate

∫3x

x3 − 3x+ 2dx

Factor the denominator gives:

∫3x

x3 − 3x+ 2dx =

∫3x

(x− 1)2(x+ 2)dx

We write:3x

(x− 1)2(x+ 2)=

A

x− 1+

B

(x− 1)2+

C

x+ 2

Notice that both of the non-trivial factors of (x−1)2, namely (x−1) and (x−1)2,are used in the partial fraction decomposition of the original function.

First Method:

We have three unknows, we need three equations. Setting x = 0, x = −1, x = 2,gives the three equations:

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−A + B +1

2C = 0

−1

2A +

1

4B + C = −3

4

A + B +1

4C =

3

2

Solving the system gives:

A = 2/3, B = 1, C = −2/3

Second Method:

A

x− 1+

B

(x− 1)2+

C

x+ 2=A(x− 1)(x+ 2) +B(x+ 2) + C(x− 1)2

(x− 1)2(x+ 2)

=A(x2 + x− 2) +B(x+ 2) + C(x2 − 2x+ 1)

(x− 1)2(x+ 2)

=Ax2 + Ax− 2A+Bx+ 2B + Cx2 − 2Cx+ C

(x− 1)2(x+ 2)

=(A+ C)x2 + (A+B − 2C)x+ (−2A+ 2B + C)

(x− 1)2(x+ 2)=

3x

x3 − 3x+ 2

Equating coefficients give:

A + C = 0

A + B − 2C = 3

−2A + 2B + C = 0

Solving the system of equation gives:

A = 2/3, B = 1, C = −2/3

The above integral becomes:∫3x

x3 − 3x+ 2dx =

∫2/3

x− 1+

1

(x− 1)2− 2/3

x+ 2dx

=2

3ln |x− 1| − 1

(x− 1)− 2

3ln |x+ 2|+ c

In the case when the denominator has repeated factors of quadratic term, all ofthe factors have to be accounted for. The numerator, of course, will be linear

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term(s):

Example: Evaluate

∫4x

(x− 1)(x2 + 1)2dx

We want4x

(x− 1)(x2 + 1)2=

A

x− 1+Bx+ C

x2 + 1+Dx+ E

(x2 + 1)2

First Method:

We have five unknowns, we need five values for x, we use: x = 0, x = −1, x = 2,x = −2, and x = 3

We have the following five equations:

−A + C + E = 0

−A2− 1

2B +

1

2C − 1

4D +

1

4E =

1

2

A +2

5B +

1

5C +

2

25D +

1

25E =

8

25

−1

3A − 2

5B +

1

5C − 2

25D +

1

25E =

8

751

2A +

3

10B +

1

10C +

3

100D +

1

100E =

3

50

Solving the system of equations give: A = 1, B = −1, C = −1, D = −2, E = 2.

Second Method:

We want4x

(x− 1)(x2 + 1)2=

A

x− 1+Bx+ C

x2 + 1+Dx+ E

(x2 + 1)2

=A(x2 + 1)2 + (Bx+ C)(x2 + 1)(x− 1) + (Dx+ E)(x− 1)

(x− 1)(x2 + 1)2

=A(x4 + 2x2 + 1) + (Bx+ C)(x3 − x2 + x− 1) + (Dx+ E)(x− 1)

(x− 1)(x2 + 1)2

=(Ax4 + 2Ax2 + A) + (Bx4 −Bx3 +Bx2 −Bx+ Cx3 − Cx2 + Cx− C) + (Dx2 −Dx+ Ex− E)

(x− 1)(x2 + 1)2

=(A+B)x4 + (−B + C)x3 + (2A+B − C +D)x2 + (−B + C −D + E)x+ (A− C − E)

(x− 1)(x2 + 1)2

Equating the coefficients we have the following system of equations:

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A + B = 0

−B + C = 0

2A + B − C + D = 0

−B + C − D + E = 4

A − C − E = 0

Solving the equations give: A = 1, B = −1, C = −1, D = −2, E = 2.

We now have:∫4x

(x− 1)(x2 + 1)2dx =

∫1

x− 1+−x− 1

x2 + 1+−2x+ 2

(x2 + 1)2dx

= ln |x− 1| −∫

x

x2 + 1dx−

∫1

x2 + 1dx− 2

∫x− 1

(x2 + 1)2dx

= ln |x− 1| − 1

2ln |x2 + 1| − tan−1(x)− 2

∫x

(x2 + 1)2− 1

(x2 + 1)2dx

= ln |x− 1| − 1

2ln |x2 + 1| − tan−1(x) +

1

x2 + 1+ 2

∫1

(x2 + 1)2dx

To integrate the last expression, we make a trig-substitution:

tan θ = x, then dx = sec2 θ dθ, cos θ =1√

x2 + 1, and the above becomes:∫

1

(x2 + 1)2dx =

∫cos4 θ sec2 θ dθ =

∫cos2 θ dθ

=

∫1

2(1 + cos 2θ) dθ =

1

2

(θ +

1

2sin(2θ)

)+ c

=1

2

(θ +

1

2(2 sin θ cos θ)

)+ c =

1

2(θ + sin θ cos θ) + c

=1

2

(tan−1(x) +

x√x2 + 1

· 1√x2 + 1

)+ c =

1

2

(tan−1(x) +

x

x2 + 1

)+ c

Putting all of these together:∫4x

(x− 1)(x2 + 1)2dx =

∫1

x− 1+−x− 1

x2 + 1+−2x+ 2

(x2 + 1)2dx

= ln |x− 1| − 1

2ln |x2 + 1| − tan−1(x) +

1

x2 + 1+ 2

∫1

(x2 + 1)2dx

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= ln |x− 1| − 1

2ln |x2 + 1| − tan−1(x) +

1

x2 + 1+ 2

(1

2

(tan−1(x) +

x

x2 + 1

))+ c

= ln |x− 1| − 1

2ln |x2 + 1| − tan−1(x) +

1

x2 + 1+ tan−1(x) +

x

x2 + 1+ c

= ln |x− 1| − 1

2ln |x2 + 1|+ x+ 1

x2 + 1+ c

Summary of Methods of Integration:

As you saw from all the integration methods we introduced, finding anti-derivativeis a much more challenging task than that of differentiation. Unlike differentia-tion, there is no straight-forward formula/rule we can use that would solve anyintegration problem. Anytime you encounter an integral, you will need to con-sider all the methods you have learned, and see which of the method(s) you canuse to solve the problem. More often than not, you will need to combine manyof the methods we covered in order to solve the problem. You may need to firstmake a substitution to change an expression to a different form, after which youmay need to integrate by parts, or you may need to make a trigonometric substi-tution then integrate using trigonometric identities. In actual problems, you willnot be told which method(s) to use, it is up to you to just keep trying differentthings, and use your creativity to come up with something that works.

You may also want to memorize some of the integration formulas that we useoften. Some of those formulas are general enough and will allow you to tackle theintegral of a collection of expressions of a certain type. Even if you do not want tomemorize too many integration formulas, at least know that there are integrationtables for the types of integrals you want to deal with. For example, you may not

want to memorize the integration formula for

∫cosn x dx, but simply knowing

that there is such a formula will let you know that cosn x can be integrated, andyou can look up the formula when needed (and allowed).

Even with the integration methods we covered, there are still many other methodsof integration that is not covered here. Moreover, there are continuous functionscomposed of elementary (polynomials, rational, radical, exponential, logarith-mic, trigonometric) functions but whose anti-derivatives cannot be expressed ascomposition of elementary functions.

Practice makes perfect. The more integration problems you do, the more youdevelop a feel for the problem and, hopefully, the more confidence you gain andwould be able to do better.

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Numerical Integration:

Because definite integrals are so useful, we want to be able to evaluate definiteintegral of all kinds; but as we saw, finding anti-derivative is not a easy task.

In the case we are interested in the value of

∫ b

a

f(x) dx, but cannot find the

anti-derivative of f(x), we can still approximate the value of the definite integralusing numerical methods.

Remember that a definite integral,

∫ b

a

f(x) dx is defined by a Riemann sum, so

we could approximate the definite integral by using a Riemann sum with largevalues of n. We could use the left endpoint or right endpoint of the partitionin the Riemann sum, or we could use the midpoint for the approximation of thedefinite integral. There are also some other formulas invented that are popular:

Midpoint Rule:

If we wanted to approximate the value of the definite integral

∫ b

a

f(x) dx, we can

divide the interval [a, b] into n equal subintervals of ∆x, and use the midpointof each subinterval, mi, as the height of the rectangle. The area of the i − threctangle will then be ∆x(f(mi)), where

mi =1

2(xi−1 + xi) is the midpoint of each interval. We have the Midpoint Rule

for approximating the value of a definite integral:

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∫ b

a

f(x) dx ≈ ∆xn∑

i=1

f(mi)

where ∆x =b− an

is the length of each interval and

mi =1

2(xi−1 + xi) is the midpoint of each interval.

In using the midpoint rules, we are approximating the definite integral (are) byusing rectangles of equal length, and the height of the rectangle is given by they−coordinate of the function at the midpoint of the interval.

Trapezoidal Rule:

f(x)

x0 x1 x2 x3 x4 x5 x6

f(x0)

f(x1)

f(x2)

f(x3)

f(x4) f(x5)

f(x6)

∆x

We will approximate a definite integral by using trapazoids instead of rectangles.Remember the area of a trapazoid with bases b1 and b2 and height h is given by

A =(b1 + b2)h

2In the picture above, the height of each trapazoid is ∆x, the two bases of thetrapazoid is f(xi−1) and f(xi), so the area of the each trapazoid Ti is

T1 =[f(x0) + f(x1)][∆x]

2

T2 =[f(x1) + f(x2)][∆x]

2

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Ti =[f(xi−1) + f(xi)][∆x]

2The sum of all these trapazoids is:

[f(x0) + f(x1)][∆x]

2+

[f(x1) + f(x2)][∆x]

2+ · · ·

+[f(xn−2) + f(xn−1)][∆x]

2+

[f(xn−1) + f(xn)][∆x]

2

=∆x

2[f(x0) + 2f(x1) + 2f(x2) + · · ·+ 2f(xn−1) + f(xn)]

This is the Trapazoidal Rule used to approximate the value of a definite inte-gral:∫ b

a

f(x) dx ≈ ∆x

2[f(x0) + 2f(x1) + 2f(x2) + · · ·+ 2f(xn−1) + f(xn)]

where ∆x =b− an

is the length of each interval.

Note that in the trapezoidal rule, the coefficient of the first and last term of thesum (f(x0) = f(a) and f(xn) = f(b) are both 1, and the coefficient of all theother terms is 2.

In using the trapezoidal rule, we are approximating the area using trapazoids ofequal altitude.

Simpson’s Rule:

We may also try to estimate a definite integral by dividing the closed interval [a, b]

into equal sub-intervals of length ∆x =b− an

. This time, we use an even number

n, and we approximate the area of the region in each sub-interval [xi−1, xi+1] byusing a parabola (quadratic function).

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x0 x1 x2 x3 x4 x5 x6

f(x0)

f(x1) f(x2)

f(x3)

f(x4)

f(x5)

f(x6)

In picture above, each xi is equally spaced. The area of the region between theinterval [x0, x2] is approximated with a parabola which contains the three pointsf(x0), f(x1), f(x2) (Three points uniquely define a quadratic function). The areaof the region between the interval [x2, x4] is approximated with a parabola thatcontains the three points f(x2), f(x3), f(x4).

We want to derive a formula for the sum of the area of these regions. To simplifythe derivation, we consider the case where x0 = −h, x1 = 0, and x1 = h.

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x0 = −h 0 h = x2

(−h, y0)

(0, y1) (h, y2)

Let (−h, y0), (0, y1), and (h, y2) be the three points that the parabola contains.Let y = Ax2 +Bx+C be the equation of the parabola that contains these threepoints, then the area under this parabola is given by:∫ h

−h

(Ax2 +Bx+ C

)dx

Using properties of integral and simplify gives:∫ h

−h

(Ax2 +Bx+ C

)dx

=

∫ h

−h

(Ax2 + C

)dx+

∫ h

−hBx dx

Notice that Bx is an odd function, so

∫ h

−hBx dx = 0, and Ax2 and C are both

even functions, so∫ h

−hAx2 + C dx = 2

∫ h

0

Ax2 + C dx.

The above simplifies to:∫ h

−h

(Ax2 +Bx+ C

)dx =

∫ h

−h

(Ax2 + C

)dx+

∫ h

−hBx dx

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= 2

∫ h

0

(Ax2 + C

)dx = 2

[A · x

3

3+ Cx

]h0

= 2

(A · h

3

3+ Ch

)=

2Ah3

3+ 2CH =

h

3

(2Ah2 + 6C

)We now want to write the expression 2Ah2 + 6C in terms of y0, y1, and y2.

Since the parabola contains the three points (−h, y0), (0, y1), (h, y2), we have:

y0 = A(−h)2 +B(−h) + C = Ah2 −Bh+ C

y1 = C ⇒ 4y1 = 4C

y2 = Ah2 +Bh+ C

So y0 + 4y1 + y2 =(Ah2 −Bh+ C

)+ (4C) + (Ah2 +Bh+ C) = 2Ah2 + 6C

So the expression for the area becomes:

h

3

(2Ah2 + 6C

)=h

3(y0 + 4y1 + y2)

The area of the region under the parabola is not changed if we shift the xito the left or right, so this formula will be valid even if we change the threex−coordinates to any arbitrary x0, x1, x2 with the same relationship. In otherwords, the area under the parabola containing the three points (x0, y0), (x1, y1),

(x2, y2) ish

3(y0 + 4y1 + y2). Similarly, the area under the parabola containing

the three points (x2, y2), (x3, y3), (x4, y4), is given by:h

3(y2 + 4y3 + y4)

We can now approximate the area, therefore the definite integral, by the area ofthese regions under the parabolas:∫ b

a

f(x) dx ≈ h

3(y0 + 4y1 + y2) +

h

3(y2 + 4y3 + y4) +

h

3(y4 + 4y5 + y6) + · · ·

+h

3(yn−4 + 4yn−3 + yn−2) +

h

3(yn−2 + 4yn−1 + yn)

=h

3(y0 + 4y1 + 2y2 + 4y3 + 2y4 + 4y5 + 2y6 + · · ·+ 2yn−2 + 4yn−1 + yn)

Letting f(xi) = yi, we have the Simpson’s Rule for estimating the value of adefinite integral:∫ b

a

f(x) dx ≈ ∆x

3[f(x0) + 4f(x1) + 2f(x2) + 4f(x3) + · · ·+ 2f(xn−2) + 4f(xn−1) + f(xn)]

where n must be an even number, ∆x =b− an

Note that the pattern in Simpson’s rule goes 1, 4, 2, 4, 2, 4, 2, . . . , 4, 2, 4, 1. Youmay want to memorize this as the coefficient of the first and last terms are 1,then the coefficients of the inner terms alternate between 4 and 2, but must start

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and end with 4.

In using Simpson’s rule, we are approximating the area using quadratic functions(parabolas) that contain the midpoint and left and right end points of the interval.

Compared to the trapezoidal rule, Simpson’s rule is more complicated to applybut it also gives a better approximation for the same n.

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Improper Integrals:

Consider the integral:∫ ∞1

1

x2dx

We can interpret this integral as representing the area under the curve of1

x2from

1 to ∞. Notice that for any real number r, the integral:∫ r

1

1

x2dx = −1

x

]r1

= 1− 1

r

is always less than 1 for any real number r > 1. This means that the arearepresented by the above definite integral is actually bounded. In fact, we candefine the above definite integral by a limit:∫ ∞1

1

x2dx = lim

t→∞

∫ t

1

1

x2dx = lim

t→∞−1

x

]t1

= limt→∞

1− 1

t= 1

In general, for any definite integral that involves an upper or lower limit(s) ofintegration to be the infinity symbol (which means that we want the (net) area ofthe region extending indefinitely to left or right (or both) sides of the real axis),we can define that as a limit at infinity.

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Definition of an Improper Integral (of type I):

If

∫ t

a

f(x) dx is defined in the interval [a, t] for all real number t ≥ a, we define:∫ ∞a

f(x) dx = limt→∞

∫ t

a

f(x) dx

If

∫ b

t

f(x) dx is defined in the interval [t, b] for all real number t ≤ b, we define:∫ b

−∞f(x) dx = lim

t→−∞

∫ b

t

f(x) dx

If the limit exists, we say that the integral is convergent, otherwise the integralis divergent

If both

∫ ∞a

f(x) dx and

∫ a

−∞f(x) dx are convergent, we can define:

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∫ ∞−∞

f(x) dx =

∫ a

−∞f(x) dx+

∫ ∞a

f(x) dx where a could be any real number.

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Example:

Evaluate

∫ ∞0

1

exdx

Ans: By definition,∫ ∞0

1

exdx = lim

t→∞

∫ t

0

1

exdx = lim

t→∞

∫ t

0

e−x dx = limt→∞−e−x

]t0

=

limt→∞

(−e−t − (−e0)

)= 0 + 1 = 1

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Example:

Evaluate

∫ ∞1

sin(1x

)x2

dx

Ans: By definition,∫ ∞1

sin(1x

)x2

dx = limt→∞

∫ t

1

sin(1x

)x2

dx

using the substitution u =1

x, we get

limt→∞

∫ t

1

sin(1x

)x2

dx = limt→∞

cos

(1

x

)]t1

= limt→∞

cos

(1

t

)− cos(1) = 1− cos(1) ≈ 0.4597

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Consider this integral:∫ 1

0

lnx dx

Notice that lnx is continuous on (0, 1], but is discontinuous at the lower limit ofintegration, 0. We can define this integral by a similar method:∫ 1

0

lnx dx = limt→0+

∫ 1

t

lnx dx

Using the method of integration by parts, we get:∫ 1

0

lnx dx = limt→0+

∫ 1

t

lnx dx = limt→0+

x lnx− x

]1t

= (0− 1)−(

limt→0+

[t− t ln t]

)=

−1− 0 = −1

Notice that we used L’Hospital’s Rule to evaluate the limit.

We can use a similar way to define an improper integral involving a function overan interval where the function has one or more points of discontinuity.

Definition of an Improper Integral (of type II):

In general, if f is continuous on the interval [a, b) but is discontinuous on b, wedefine:∫ b

a

f(x) dx = limt→b−

∫ t

a

f(x) dx

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Similarly, if f is continuous on (a, b] but discontinuous on a, we define:∫ b

a

f(x) dx = limt→a+

∫ b

t

f(x) dx

If there is a number c, a < c < b and f is continuous on [a, b] except c, and

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∫ c

a

f(x) dx and

∫ b

c

f(x) dx both exists, then we define∫ b

a

f(x) dx =

∫ c

a

f(x) dx+

∫ b

c

f(x) dx

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Example:

Evaluate:

∫ 2

−1x−1/3 dx

Ans: Notice that x−1/3 is continuous on [−1, 2] except at x = 0, so by definitionwe have:∫ 2

−1x−1/3 dx = lim

t→0−

∫ t

−1x−1/3 dx+ lim

t→0+

∫ 2

t

x−1/3 dx= limt→0−

3

2x2/3

]t−1

+ limt→0+

3

2x2/3

]2t

= limt→0−

(3

2t2/3 − 3

2(−1)2/3

)+ lim

t→0+

(3

2(2)2/3 − 3

2(t)2/3

)=

(0− 3

2

)+

(3

2· 3√

4− 0

)

= −3

2+

3

2· 3√

4 ≈ 0.88

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Example:

Evaluate

∫ 1

0

1

xdx

Ans: By definition,∫ 1

0

1

xdx = lim

t→0+

∫ 1

t

1

xdx = lim

t→0+ln |x|

]1t

= 0− limt→0+

ln |t| = 0− (−∞) =∞

Since ∞ is not a real number, the indefinite integral is divergent.

We know that if f(x) is an odd function and is continuous over the interval

[−a, a], then

∫ a

−af(x) dx = 0. This is not true if f(x) fails to be continuous in

the interval.

Example:∫ 1

−1

1

xdx is divergent since lim

t→0−

∫ t

−1

1

xdx is divergent. This improper integral is

not equal to 0.

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There are times when it will be difficult or even impossible to evaluate an im-proper integral, but it would still be helpful to know if the improper integral isconvergent or not. A theorem similar to the squeeze theorem can be used todetermine if some improper integral is convergent or not:

Comparison Test For Integrals:

Suppose f and g are continuous on [a,∞) and f(x) ≥ g(x) ≥ 0 for all x ≥ a:

If

∫ ∞a

f(x) dx is convergent, then

∫ ∞a

g(x) dx is also convergent.

If

∫ ∞a

g(x) dx is divergent, then

∫ ∞a

f(x) dx is also divergent.

Example:

Determine if

∫ ∞1

e−x2

dx is convergent.

Ans: It is not easy to find the anti-derivative of e−x2

. Instead, since e−x2 ≤ e−x

for all x ≥ 1, and we know that

∫ ∞1

e−x dx is convergent (why?), so by the

comparison test,

∫ ∞1

e−x2

dx must also be convergent.