Integration tipsMath 252

C. M. Hughes

October 25, 2013

We have so far studied integration in a conceptual form in terms of left and right sums, and thearea between the graph of a function and the x axis. We have used the Fundamental Theoremof Calculus to obtain anti-derivatives analytically, which as we have seen is a skill that requires athorough knowledge of differentiation.

The aim of this handout is to clarify the ideas and tools that we have dealt with so far. This handouthas been divided into categories that increase in complexity; we begin with the simple cases. Weshall assume throughout this handout that C is a constant of integration.

Anti-derivatives: simple cases

1 By far the most simple case that we can ever hope to come across are the power functionsand polynomials. We know that when we differentiation a power function, we simply useThe Power Rule

ddx

xn = nxn1,

where n is any real number. Reversing the power rule to find an anti-derivative is equallystraight forward, and we have that

xndx =xn+1

n+ 1+ C ,

where n is any real number except n = 1 (we shall deal with this case in due course). Wecan always check our anti-derivatives by differentiating them- if we do not obtain the correctresult, then we need to review our answer (and perhaps our differentiation technique). Thisis built upon in Table 1.

2 In the case n= 1, we are considering1xd x = ln(|x |) + C , x 6= 0.

A quick mental check will show that this is the correct anti-derivative. Note the use of |x | toshow that we can not take the natural log of a negative number.

3 We have also considered anti-derivatives of exponential and trigonometric functions. In thecase where the argument of these functions is just x , or a constant multiple thereof, thenthe anti-derivative is straight forward to find- see the examples given in Table 1. Note thatfor each example, there is no product of functions, so reversing the chain rule simply comesdown to dividing by a constant.

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Table 1: Some indefinite integrals, and their anti-derivatives. Note that the argument of eachexample is never more than a constant times x , so the integration (and differentiationto check the answer) is straight forward. Verify each one by differentiating the anti-derivative.

Integral Anti-derivative Notes(3x + 2)10dx

133

(3x + 2)11 + C Note that if there were anythingmore complicated than a constantmultiple of x in the bracket, theintegration would be a lot more in-volved.

e2xdxe2x

2+ C The argument of e2x is 2x , so

when differentiating (by the chainrule), we only have to account forthe 2.

e2x+1dxe2x+1

2+ C The argument of e2x+1 is 2x+1, so

when differentiating (by the chainrule), as above we only have to ac-count for the 2.

e3x+1dxe3x+1

3+ C The argument of e3x+1 is 3x+1, so

we only need to account for the 3when differentiating (by the chainrule).

ekx+bdx (k, b constant)ekx+b

k+ C The argument of ekx+b is kx+b, so

we only need to account for the kwhen differentiating (by the chainrule).

cos(x)dx sin(x) + C The argument of cos(x) is just x ,so we dont need to account forany other constants.

sin(x)dx cos(x) + C The argument of sin(x) is just x ,so we dont need to account forany other constants.

cos(kx + b)dx (k, b constant)sin(kx + b)

k+ C The argument of cos(x) is kx + b,

so we need to account for the kwhen differentiating (by the chainrule).

Notes on Table 1In each of the indefinite integrals of Table 1, we could multiply by a constant and the re-sulting anti-derivative would simply be multiplied by the same constant. Introducing a newfunction of x within the integral sign complicates things dramatically, as does changing thearguments of ex , sin(x), etc. We examine these next.

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More complicated integrals: The method of substitution or Guess and checkThese methods rely heavily on the ability to spot a useful substitution or to make an educatedguess (respectively). Making a substitution should make the integral more manageable, andhelps to reverse the chain rule, and in fact will often work even when the chain rule is notinvolved.

4 Consider x(x2 + 3)2dx . (1)

Notice that x is nearly the derivative of x2 + 3, which is useful as we guess that the chainrule will be used here. Consider the function

(x2 + 3)3.

Then,

ddx

(x2 + 3)3 = 3(x2 + 3)2(2x)

= 6x(x2 + 3)2,

which is nearly the expression under our integral sign, but is out by a factor of 6. We thereforeuse this function as the anti-derivative, but divide by 6, so that

x(x2 + 3)2dx =(x2 + 3)3

6+ C .

If we wanted to use the method of substitution on this integral, then we make the substitution

u= x2 + 3,

so thatdu= 2xd x .

We now use this in equation (1), which saysx(x2 + 3)2dx =

(x2 + 3)2

u2xd xdu2

=

u2

2du

=u3

6+ C

=16(x2 + 3) + C .

5 Lets consider another example: x2 cos(x3)dx . (2)

The first thing we look for is a straight forward application of the chain rule- i.e, did the firstfunction come from differentiating the inside part of the second function? In this case, theanswer is yes, but with a factor of 3 that we need to take of. So lets consider

sin(x3),3

and when differentiated, this gives

ddx

sin(x3) = cos(x3)(3x2)

= 3x2 cos(x3).

This is nearly our integrand, but were out by a factor of 3. Therefore, our answer isx2 cos(x3)dx =

sin(x3)3

+ C .

If we wanted to approach this using substitution, we would make the substitution

u= x3,

as it is the inside function, so that

du= 3x2dx .

Our indefinite integral in (2) can now be written asx2 cos(x3)dx =

cos(x3) cos(u)

x2dxdu3

=

cos(u)

3du

=sin(u)

3+ C

=13

sin(x3) + C

6 In the case of the integrand being a rational function, the first thing to look for is an appli-cation of the chain rule on a function involving logarithms. Recall from Math 251, that

ddx

ln( f (x)) =f (x)f (x)

. (3)

Consider ex + 1ex + x

d x .

The first thing to notice here, is that when you differentiate the denominator, you obtainthe numerator. From the statement we wrote in (3), our instinct is to look for a logarithmicfunction.

Considerln(ex + x),

thenddx

ln(ex + x) =ex + 1ex + x

=

f (x)f (x)

So this is our answer, and

ex + 1ex + x

d x = ln (|ex + x |) + C .This question could also be done by substitution, but that is left as an exercise.

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7 The method of substitution can also be used for problems that have not arisen by differenti-ating using the chain rule, for example

x2x + 1

dx .

For this example, try u= 2x + 1.

Integration by partsWe have encountered indefinite integrals that can be integrated directly, or by substitution,and have also seen how we can integrate by parts.

We use the formula udv = uv

vdu, (4)

which is derived by integrating the product rule for differentiation. The technique relies onchoosing the functions u and v so that the problem becomes more manageable, and bear inmind that

we need to be able to differentiate u we need to be able to integrate dv we need to be able to integrate the product vdu.

Generally speaking, it will quickly become obvious if an unwise choice has been made, as theproblem will become significantly more complicated. Do not worry if you make the wrongchoice at first, experience coupled with trial and error will guide you. It is hoped that thefollowing examples will also assist you. We shall assume throughout this handout that C isa constant of integration.

8 Consider x sin(x)dx .

We notice first of all that this is not an obvious candidate for substitution as the two functionsx and sin(x) are not related by the chain rule.

Let us make the following observations about possible choices of u and dv

if we choose u= x , then du= dx , which would almost certainly simplify the next step if we choose u= x , then we would have to choose dv = sin(x)dx , which can easily be

integrated to give v = cos(x), so this seems like a good choice.Based on these comments, we have

u= x , dv = sin(x)dx ,du= dx , v = cos(x),

and therefore, using (4),x sin(x)dx = x

u( cos(x))

v

( cos(x) v

(1)du

dx

= x cos(x) +

cos(x)dx

= x cos(x) + sin(x) + C .5

Notice that with our choice of u = x , the polynomial part of the integral went away, asdu= 1. If we had chosen u= sin(x), dv = x , then

u= sin(x), dv = xd x ,

du= cos(x)dx , v = x2/2.

Using (4) with these choices would have givenx sin(x)dx =

x2

2sin(x)

x2

2cos(x)dx ,

which has clearly not improved our situation. In both choices of u and v, the trigonometricfunctions will always differentiate or integrate to give another trigonometric function.

9 Consider x2exdx .

As with the previous example, there is no obvious substitution or reversal of the chain rulethat would progress us.

We shall try and use integration by parts. Based on our findings from the previous example,lets choose

u= x2, dv = exdx ,du= 2xd x v = ex .

Observe that v and dv are exactly the same (properties of the exponential function). Reduc-ing the power of polynomials by choosing u as the polynomial function when integrating byparts is a often a good choice.

So, with the above choices of u and dv, we havex2exdx = x2ex

2xexdx ,

from which we only have to worry about the second integral. Given our experiences so far,we notice that we can use integration by parts on this integral with the following choices ofu and v

u= 2x , dv = exdx ,du= 2d x , v = ex .

This now gives x2exdx = x2ex

2xex

2exdx

= x2ex 2xex + 2ex + C= (x2 2x + 2)ex + C .

This example was slightly more involved as we had to use integration by parts twice. Noticethat no matter what our choices of u and v, we would not have been able to change theexponential term, no matter how many terms we integrate or differentiate.

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10 Evaluating definite integrals using integration by parts is done in the following way. Weconsider pi

0

x sin(x)dx .

Then from our work earlier on this handout, we have that pi0

x sin(x)dx = [x cos(x)]pi0 + pi

0

cos(x)dx

= pi cos(pi) (0 cos(0)) + [sin(x)]pi0= pi(1) 0+ (0 0)= pi

Trigonometric functionsThe following three identities (among others) are often useful when dealing with trigono-metric functions sin

2( ) + cos2( )=1,sin(m n )=sin(m ) cos(n ) sin(n ) cos(m ),cos(m n )=cos(m ) cos(n ) sin(m ) sin(n ).

11 These identities can be used together in the case m= n= 1 to give

cos(2 ) = cos2( ) sin2( )= 1 sin2( ) sin2( )= 1 2 sin2( ),

and therefore

sin2( ) =1 cos(2 )

2.

This provides an alternative (and easier) way of finding

sin2( )d . Similarly, we have

cos2( ) =cos(2 ) 1

2.

12 These identities give

sin(m + n ) = sin(m ) cos(n ) + sin(n ) cos(m ), (5)sin(m n ) = sin(m ) cos(n ) sin(n ) cos(m ). (6)

Adding equations (5) and (6) gives

sin(m ) cos(n ) =12

sin((m+ n) ) + sin((m n) )

.

This gives us a tidy way of findingsin(m ) cos(n )d = 1

2

cos((m+ n)

m+ n+

cos((m n) )m n

,

but with the restriction m 6= n.7

13 Similarly, we can show thatcos(m ) cos(n )d =

12

sin((m+ n) )

m+ n+

sin((m n) )m n

,

sin(m ) sin(n )d =12

sin((m n) )

m n +sin((m+ n) )

m+ n

,

both in the case m 6= n.Questions to ask when integrating

1. Is the integrand a simple function (e.g. ex , sin(x)), which has an argument that is nomore complicated than kx + b, where k and b are constant?

Yes: the anti-derivative should be straight forward, but remember to account forany constants that arise from the chain rule (see Table 1)

No: try the next question.2. Is the integrand a product of functions?

Yes: can you spot a good substitution, or a guess and check? No: try integration by parts:

(a) We need to choose u and dv so that we

can differentiate u

can integrate dv

can integrate the product vdu.

If it looks more complicated than your original problem, consider changingyour choice, but also be prepared to persevere.

(b) If the integrand involves a polynomial or power function, remember that dif-ferentiating it will often be a good choice, but not always

(c) cos(x), sin(x), and ex will always remain in the problem, no matter how manytimes they are integrated or differentiated

(d) You may need to use integration by parts more than once

(e) You may get cases when you have to rearrange in terms of the original integral(see our notes on

sin2(x)dx for example).

3. Is the integrand a rational function that has arisen from a natural logarithm function?

Yes: try a logarithmic function, but be careful to account for any constants No: there may still be a substitution that will improve things, or in the case of

polynomial functions, try partial fractions.

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