Chapter 10
CHAPTER 9 SOLUTIONS TO REINFORCEMENT EXERCISES IN TECHNIQUES OF INTEGRATION
9.3.1 Definition of integration
9.3.1A.
Differentiate the following functions
i) 3x3 ii) iii) iv) x
v) sin 3x vi) cos x3 vii) e5x viii)
ix) ln 2x x) xi) ln(3x + 1) xii) tan–1x
Solution
All these are straightforward differentiations, designed to prepare you up for integration!
i) = 9x2
ii) = = x1/2 – 1 = x– 1/2 =
iii) = 2 = – 8x – 5 = –
iv) = x2/5 – 1 = x– 3/5
v) = 3 cos 3x by function of a function rule
vi) = – 3x2 sinby function of a function rule
vii) = 5e5x by function of a function rule
viii) = = – (x + 1)– 2
= –
ix) = =
x) = = – 1/2
=
xi) =
=
xii) =
9.3.1B.
Integrate the following functions
i) ii) 3e5x iii) iv)
v) 3x2 cos x3 vi) vii) x viii)
ix) 2 cos 3x x) x2 xi) xii)
Solution
All these integrals come from the results of A – but in a different order! C is the ubiquitous arbitrary constant of course.
i) is deliberately tricky, to encourage care when simply reversing derivatives. From A ix) we have
=
which would lead us to
dx = ln 2x + C
whereas we know from the standard integral that
dx = ln x + C
But of course there is no real problem because we can write
ln 2x + C = ln x + ln 2 + C
and then simply absorb the ln 2 in the arbitrary constant. So, using the standard integral we have
dx = 2 dx = 2 ln x + C
ii) For 3e5x compare A vii):
= 5e5x
So
5e5x dx = e5x + C
Therefore
3e5x dx = 5e5x dx = e5x + C
iii) Comparing with A viii) we have
dx = – 2 dx = – + C
iv) Comparing with A iii) we have
dx = – dx = – + C = – + C
v) A vi) gave us
= – 3x2 sin
but we want to integrate 3x2 cos and this suggests we swap round sin and cos:
= 3x2 cos
and so
3x2 cosdx = sin+ C
vi) From A xii)
dx = – 3dx
= – 3 tan– 1 x + C
vii) From A iv)
xdx = xdx = x+ C
viii) From A ii)
dx = 2 dx = 2 x– dx = 2 + C
ix) From A v)
2 cos 3x dx = 3 cos 3x dx
= sin 3x + C
x) From A i)
x2 dx = 9 x2 dx =
= x3 + C
which, by now, you should see directly anyway.
xi) From A x)
dx = 6 dx = 6 + C
xii) From A xi)
dx = dx = ln (3x + 1) + C
9.3.2. Standard integrals
9.3.2A.
Integrate the following functions
i) 4 ii) 3x2 – 2x + 1 iii) 3/x5 iv) x
v) cosx vi) sec2x vii) lnx viii)
Solution
All of these rely on the standard integrals in UEM Table 9.1 and very simple properties of integrals.
i) Here we can take the 4 out of the integral to give
4 dx = 4 dx = 4x + C
ii) In integrating any polynomial it is basically integrated as a sum of power functions
dx = x3 – x2 + x + C
ii) Such things as inverse powers and roots are best put in power form before using standard integrals
3/x5 dx = 3 x– 5 dx = – x– 4 + C
iv)xdx = x+ C
v) cosx dx = sin x + C
vi) sec2x dx = tan x + C
vii) ln x dx = x ln x – x + C
While this integral is not in Table 9.1, it is normally regarded as a ‘standard integral’ because it is the integral of one of the elementary functions. We see how to actually integrate this in UEM Section 9.2.10.
viii)dx = tan– 1 x + C
9.3.2B.
Integrate the following functions with respect to the appropriate variables
i) 4u ii) 2s2 – 3s + 2 iii) 6/t7 iv) x
v) sin θ vi) cosec2 t vii) e3t viii)
Solution
i) 4u du = 4 u du = 4+ C = 2u2 + C
ii) ds = – + 2s + C
iii) dt = 6 t– 7 dt = t– 7 + 1 + C = – + C
iv)xdx = + C = x+ C
v) sin θ dθ = – cos θ + C
vi) cosec2 t dt = – cot t + C
vii) e3t dt = + C
viii) ds = tan– 1 s + C
9.3.3 Addition of integrals
Integrate
i) 2x2 – ii) 2 cos x – sin x iii) cosh x =
iv) sin(x + 3) by compound angle formulae
v) The polynomial:-
arxr
vi) sin x cos 2x vii) sinh x
Solution
i) dx = 2 dx – 3 dx
= x3 + + C
ii) dx = 2cos x dx – sin x dx = 2 sin x + cos x + C
iii) cosh x dx = dx = ex dx + e– x dx
= ex – e– x + C = sinh x + C
iv) sin(x + 3) dx = dx
= cos 3sin x dx + sin 3 cos x dx
= – cos 3 cos x + sin 3 sin x + C
= – (cos x cos 3 – sinx sin 3) + C
= – cos(x + 3) + C
v) arxr dx = arxr dx = ar + C = + C
Note that we can combine the arbitrary constants added to each term of the sum into one summed arbitrary constant.
vi) We can use the compound angle formulae to express products of sine and cosine in terms of sums and differences of sines and cosines:
sin 3x = sin(2x + x) = sin 2x cos x + sin x cos 2x
sin x = sin(2x – x) = sin 2x cos x – sin x cos 2x
Subtracting gives
sin x cos 2x =
So
sin x cos 2x dx = dx
= + C
vii) sinh x dx = dx = ex dx – e– x dx
= ex + e– x + C = cosh x + C
9.3.4 Simplifying the integrand
9.3.4A.
Evaluate the following by simplifying the integrand – noting any special conditions needed.
i) dx ii) lnecosx dx
iii) dx iv) dx
v) 2 dx
Solution
i) Here it is a matter of factoring the numerator and cancelling down
dx = dx = dx = – x + C
provided that x ( 2.
ii) From properties of logs, we get lnecosx = cos x, and so
lnecosx dx = cos x dx = sin x + C
iii) Remembering (hopefully!) that
cos 2x = cos2 x – sin2 x = (cos x – sin x)(cos x + sin x)
from the difference of two squares, we get
dx = dx
= dx
= sin x + cos x + C
provided of course that cos x – sin x ( 0.
iv) dx looks a very messy animal and we have to take it a step at a time. We notice from the properties of logs and exponentials that
lne2x = 2x
e3lnx = elnx3 = x3
= ecos2x e+ sin2x = ecos2x+ sin2x = e1 = e
and so
= =
Now we have
dx = dx = – + C
In this case we must have x > 0 for ln x3 to exist.
v) Once again we must call upon our trig identities
2 dx = dx =
dx = x + cos 2x + C
9.3.4B.
Sometimes you have to complicate a function to integrate it. Integrate sec x by multiplying and dividing by sec x + tan x. Integrate cosec x in a similar way.
Solution
This may be quite a tough question for you at this stage, and you may want to come back to it after Section 10.6 – it has more of a novelty value here. As instructed, we have
sec x =
You may notice that the top is now the derivative of the bottom:
= sec x tan x + sec2 x
= sec x(sec x + tan x)
So if we make the substitution u = sec x + tan x (Section 10.6) we get
du = sec x(sec x + tan x) dx
and so
sec x dx = dx = = ln u + C
= ln(sec x + tan x) + C
We can use exactly the same trick with cosec x, which you should try now. The result is
cosec x dx = ln (cosec x – cot x) + C
9.3.5 Linear substitution in integration
Integrate
i) (4x + 3)7 ii) (2x – 1)iii) sin(3x – 1) iv) 2e4x + 1
v) vi) (2x – 1)4 vii)viii) cos(2x + 1)
ix) 4e3x–1 x)
Solution
i) To integrate (4x + 3)7 we put u = 4x + 3. Then du = 4 dx and we get
7 dx = u7 = u8 = u8
= 8 + C
Try to aim for a facility such that you don't have to actually substitute the u variable:
7 dx = 7 d(4x + 3)
= 8
= 8 + C
ii) In this case substitute u = 2x – 1 to get
dx = u1/3 = u4/3 = u4/3
= 4/3 + C
iii) With u = 3x – 1 we have
sin(3x –1) dx = sin u = – cos u = – cos (3x – 1) + C
iv) Using u = 4x you can confirm that
dx = + x + C
= e4x + x + C
v) Substituting u = 2x + 1 gives
dx = ln (2x + 1) + C
vi) With u = 2x – 1
4 dx = 5 = 5 + C
vii) u = 3x + 2
dx = )dx = + C
= + C
viii) u = 2x + 1
cos(2x + 1) dx = sin(2x + 1) + C
ix) u = 3x – 1
4e3x–1 dx = 4 e3x–1 + C = e3x–1 + C
x) u = 4x – 3
dx = ln (4x – 3) + C
9.3.6 The du = f´(x) dx substitution
9.3.6A.
Write down the substitution u = f(x) you would use to integrate the following and hence integrate them.
i) x2ex3 ii) sec2x sin(tan x + 2)
iii) cos x sin3x iv) v) tanx
Solution
i) In the case of the integrand x2ex3 we notice that the derivative of the x3 in the exponential is 3x2 which is, apart from the constant 3, what we have as the other factor in the integrand. This suggests that we substitute u = x3. Then du = 3x2 dx and we get
x2ex3dx = ex3 x2dx = ex3 3x2dx
= eu du = eu + C
= ex3 + C
ii) In sec2x sin(tan x + 2) noting that sec2x is the derivative of tan x + 2 points to the substitution u = tan x + 2. Then du = sec2x dx and
sec2x sin(tan x + 2) dx = sin(tan x + 2) (sec2x dx)
= sin u du = – cos u + C
= – cos(tan x + 2) + C
iii) cos x sin3x = cos x (sin x)3 and of course the derivative of sin x is cos x. By the (hopefully) now familiar process we put u = sin x and obtain
cos x sin3x dx = u3 du = + C
= + C
iv) With u= x2 + 2x + 3, du = (2x + 2)dx = 2(x + 1)dx, so
dx = dx
= = ln u + C
= ln(x2 + 2x + 3) + C
v) Remembering tanx = and noting that – sin x is the derivative of cos x we have, putting u = cos x
tan x dx = dx
= – = – ln u + C
= – ln(cos x) + C = ln + C
= ln (sec x) + C
9.3.6B.
Integrate
i) f´(x) sin(f(x)) ii) f(x)e(f(x))2 f´(x) iii) x2 cos
iv) – sin x ln(cos x) v) vi) f´(x)ef(x) dx
vii) f´(x) cos(f(x) + 2) dx
Solution
In these questions we are basically generalizing what we did in Question A
i) In f´(x) sin(f(x)) we put u = f(x), du = f´(x) dx to obtain
f´(x) sin(f(x)) dx = sin u du
= – cos u + C
= – cos (f(x)) + C
ii) We put u = f(x), du = f´(x) dx to obtain
f(x)e(f(x))2 f´(x) dx = u eu2 du
= eu2 + C
= ef(x)2 + C
iii) Perhaps you are now getting the hang of this method of substitution. Here we give the sort of approach you might follow if you are really proficient
x2 cos dx = cos 3x2 dx
= cos d(x3 + 1)
= sin + C
iv) Again, let us try to move quite quickly
– sin x ln(cos x) dx = ln (cos x) d(cos x)
= cos x ln (cos x) – cos x + C
v) This one is designed to discourage you from getting stuck on tramlines! In fact we don't need to substitute in this one:
dx = dx = dx
= ln (x – 1) + C
vi) Putting u = f(x), du = f´(x)dx gives
f´(x)ef(x) dx = ef(x) (f´(x)dx) = eu du
= eu + C = ef(x) + C
vii) f´(x) cos(f(x) + 2) dx = cos(f(x) + 2) (f´(x)dx)
= cos(f(x) + 2) ((f(x) + 2)′dx)
= sin (f(x) + 2) + C
9.3.6C.
Integrate
i) 3x23 ii) 2x iii)
iv) (x + 1)cos v) sin x ecos x vi)
vii) cos 2x(2 sin 2x + 1)3 viii)
Solution
In all of these we look for a function and its derivative. We will do the first in detail and then simply give the substitutions for the rest.
i) In 3x23 we note that 3x2 is the derivative of x3 + 2 so we make the substitution u = x3 + 2 du = 3x2dx and so
3x23 dx = 3 d
Now put u = x3 + 2:
=u3 du = + C
= + C
ii) 2xdx = = d(x2 + 1)
= 3/2 + C
iii) dx = = ln+ C
iv) cosdx = cosd
= sin+ C
v) sin x ecos x dx = – ecos x d(cos x)
= – ecos x + C
vi) dx = = ln (tan x) + C
vii) cos 2x(2 sin 2x + 1)3 dx = 3 4cos 2x dx
= 3 d(2 sin 2x + 1)
= + C = + C
viii) dx = 3dx = 3
= 6 + C
9.3.7 Integrating rational functions
9.3.7A.
Integrate the following by partial fractions
i) ii) iii)
iv) v) vi)
Solution
i) dx = dx
= dx + dx
= ln (x – 1) + ln (x + 3) + C
= ln + C
ii) dx = dx
= dx
= – ln (x + 2) + 2 ln (x + 3) + C
= ln + C
iii) dx = 4
= dx
= + C
= ln + C
= ln + C
iv) In this case the partial fractions is more complicated. We write
∫ +
= + by the cover-up rule
=
So equating numerators we get
3(x2 + 1) + 2(x + 1)(Bx + C) ∫ 3
The coefficient of x2 then gives B = – and putting x = 0 gives C =
So
∫ –
and therefore
dx = – dx
= ln (x + 1) – dx +
= = ln (x + 1) – ln (x2 + 1) + tan(1 x + C
=ln+ tan– 1 x + C
v) In this case you can confirm that
= – –
So
dx = 3 – 3 – 2
= 3 ln(x – 2) – 3 ln(x – 1) + + C
= 3 ln+ + C
vi) dx = dx
+ –
= ln(x – 1) + ln(x + 1) – ln(x + 2) + C
= ln+ C
= ln+ C
9.3.7B.
Integrate the following by completing the square
i) ii) iii)
iv) v)
Solution
i) By completing the square we have
x2 + 2x + 5 = (x + 1)2 + 4
and so
dx =
= dx = d(x + 1)
= tan– 1 + C
ii) dx = dx = 3 tan– 1 (x – 1) + C
iii) dx = 4
= 2 = 2 tan– 1 (2x + 1) + C
iv) dx = = tan– 1 (x + 3) + C
v) Note that 2x2 + 12x + 27 = 2(x2 + 6x) + 27 = 2((x + 3)2 – 9) + 27
= 2(x + 3)2 + 9
So
=
=
= tan– 1 + C
9.3.7C.
Integrate
i) ii) iii)
Solution
In this question we have to divide out first.
i) First note that by long division
=
= 1 + = 1 +
= 1 + + by partial fractions
So
dx = dx
= x + ln(x – 2) – ln(x + 1) + C
= x + ln||+ C
ii) Even more messy algebra here, but basically the same idea. On long division we get
= x – 2 +
So
dx = dx
= – 2x + 2 dx
= – 2x + dx
(here setting up the integrand to split into two parts more easily dealt with)
– 2x + + 2
= – 2x + ln(x2 + 2x + 2) + 2 tan– 1 (x + 1) + C
by a substitution and a completing the square
iii) This one is more of a test of your algebra, for which you really need your wits about you to avoid some nasty calculations! we have
=
= = 3x2 +
= 3x2 + 3 +
So
dx = dx
= x3 + 3x + 3 dx
= x3 + 3x + ln|| + C
9.3.8 Using trig identities in integration
Integrate
i) cos2 x sin3 x ii) cos2x cos3x iii) cos5x
iv) cos 5x sin 3x v) sin 2x sin 3x
Solution
i) In a case like cos2 x sin3 x we remember that – sinx is the derivative of cos x and try to express everything in terms of cos x and it's derivative. With this in mind we have
cos2 x sin3 x dx = cos2 x sin2 x sin x dx
= cos2 x (1 – cos2 x)sin x dx = – cos2 x (1 – cos2 x) d(cos x)
= d(cos x) = – + C
ii) For cos2x cos3x we use compound angle formulae to obtain
cos2x cos3x dx = dx
= dx
(remember cos(– x) = cos x)
= sin 5x + sin x + C
iii) Similar to i)
cos5x dx = cos4 x cos x dx
= 2 d(sin x) = d(sin x)
= sin x – sin3 x + sin5 x + C
iv) Just like ii)
cos 5x sin 3x dx = dx
= – + C
= – + C
v) sin 2x sin 3x dx = dx
= + C
9.3.9 Using trig substitutions in integration
9.3.9A.
Integrate the following functions using appropriate substitutions
i) ii) iii)
iv) v) vi)
Solution
The most we need to do to get these into standard integral form is complete the square, aiming for the standard integrals (Table 9.1)
= sin– 1
dx = tan– 1
i) = = sin– 1 x + C
ii) dx = dx
= tan– 1 + C
iii) dx = = sin– 1 (3x) + C
iv) dx = = tan– 1 (2x) + C
v) Here we need to complete the square. Noting that
8 – 2x – x2 = 32 – (x + 1)2
we have
= = sin– 1 + C
vi) = = sin– 1 + C
9.3.9B.
Use the t = tan substitution to integrate dθ
Solution
With t = tan then dθ = , cos θ =
So
dθ =
= = )
= dt
=
= ln + C = ln|| + C
9.3.10 Integration by parts
Integrate
i) x cos x ii) x3ex iii) sin–1 x iv) ex cos x
v) x2 cos x vi) x ln x vii) x3ex2
Solution
i) Start by integrating the cos x
x cos x dx = x sin x – sin x dx = x sin x + cos x + C
ii) Integrate the ex first, and keep on with it!
x3ex dx = x3ex – 3 x2ex dx
= x3ex – 3 = x3ex – 3 x2ex + 6
= x3ex – 3 x2ex + 6 xex – 6ex + C
= ex + C
iii) In the case of sin–1 x it is clear that this is similar to ln x in that we will need to differentiate the sin–1 x and therefore integrate the invisible 1 that is with it. So
sin–1 x dx = 1 sin–1 x dx
= x sin–1 x –
= x sin–1 x – + C
where the last integral is done by substituting u =
iv) Here we go round in circles
I = ex cos x dx = ex cos x – ex (– sin x) dx
= ex cos x + ex sin x dx = ex cos x + ex sin x – ex cos x dx
= ex cos x + ex sin x + C – I
So, solving for I
I = + C
v) Just keep differentiating the x2 until it goes away!
x2 cos x dx = x2 sin x – 2x sin x dx
x2 sin x – 2
= x2 sin x + 2x cos x – 2 sin x + C
vi) x ln x dx = ln x – dx = ln x – dx
= ln x – + C
vii) Note that we can integrate xex2 but not ex2, and so
x3ex2 dx = x2 (xex2)dx = x2 ex2 – 2x ex2 dx = x2 ex2 – ex2 + C
= ex2 + C
9.3.11 Choice of integration methods
9.3.11A.
From the following three methods of integration:
A standard integral
B substitution
C integration by parts
state which you would use to evaluate the following integrals :-
i) 2ex dx ii) x ex dx iii) x ex2 dx
iv)x cos dx v) sec x tan x dx vi) sec2x tan x dx
Solution
Note that you are not asked to integrate these integrals - although no one is stopping you!
i) 2ex dx = 2 ex dx is simply a standard integral, so method A
ii) x ex dx is a standard integration by parts, so method C
iii) x ex2 dx might look like integration by parts, because it is a product, but in fact the method here is B, using a substitution u = x2
iv)x cos dx is again B, using a substitution u = x2
v) sec x tan x dx is a standard integral (giving sec x), so A
vi) By remembering that the derivative of tan x is sec2x we see that sec2x tan x dx will succumb to a substitution u = tan x, so method B
9.3.11B.
Choose from
A partial fractions
B completing the square
C substitution
the methods (in the order in which you would use them) you would use to integrate
i) ii) iii)
iv) v) vi)
Solution
i) In the denominator doesn't factorize so the only option is completing the square:
=
Then use substition u = x + to convert to a standard integral of the form
dx = tan– 1
So B, C
ii) The denominator factorizes
=
So use partial fractions and substitutions in the resulting reciprocal integrals. So A, C
iii) In the numerator is the deriviative of the denominator, so use the substitution u = x2 + x – 1. So C.
iv) Noting that 4x + 12 is twice the derivative of x2 + 6x + 6 we again use substitution u = x2 + 6x + 6 to integrate . So C.
v) The only possibility with is to complete the square:
=
and then substitute u = x + 3 to get an integral of the form
= sin– 1
So B, C
vi) = is a simple example of completing the square and then substitution u = x + 1. So B, C
9.3.11C.
Integrate the integrals in Review Question 9.1.11
Solution
This question is designed to pull together lots of different integration methods and properties of functions. You should have already decided how to tackle each one and here we simply go ahead and integrate them.
i) dx == ln (x2 – 6x + 4) + C
ii) sin3x dx = sin2x sin x dx
= sin x dx = – d(cos x)
= – cos x + C
iii) dx = dx = 1dx = x + C
iv) dx = dx =
= ln(2x + 3) + C
v) x e3x2 dx = e3x2 6xdx = e3x2 d(3x2) = e3x2 + C
vi) dx = dx = 3 sin– 1+ C
vii) x sin(x + 1) dx = – x cos(x + 1) +cos(x + 1) dx
= – x cos(x + 1) + sin(x + 1) + C
viii) cos4x dx = 2 dx = 2 dx
= dx = x + sin 2x + dx
= x + sin 2x + sin 4x + C
ix) ln() dx = dx = – x ln x + x + C
x) dx = dx = dx
= 5 – 4 = 5 ln(x – 3) – 4 ln(x – 2) + C
xi) x e2x dx = – e2x dx = – e2x + C
= e2x(2x – 1) + C
xii) ex cos (ex ) dx = cos (ex ) d(ex) = sin(ex) + C
xiii) sin 2x cos 2x dx = sin 4x dx = – cos 4x + C
xiv) dx = = + C
xv) dx = dx + 2
= ln(x2 + 2x + 2) + 2
= ln(x2 + 2x + 2) + 2 tan– 1(x + 1) + C
xvi)sin 4x cos 5x dx = dx
= dx= – cos 9x + cos x + C
= + C
xvii) x cos(x2 + 1) dx = cos(x2 + 1) d(x2 + 1)
= sin(x2 + 1) + C
9.3.12 The definite integral
9.3.12A.
Evaluate
i) x2 dx ii) x cos x dx iii) dx
iv) dx v) xex dx
Solution
i) x2 dx = =
ii) When using limits in integration by parts we can substitute the limits in as we go along. So
x cos x dx = – sin x dx
= + = – 1
NB Note that the angle must be given as radians when it arises from integration in this way – it should not be 90( – 1. It is always best to use radians for angular measure in any use of calculus.
iii) For dx we need a substitution, and in doing this we must change the integration limits to those appropriate to the new variable. The substitution here is u = x3 + 1, then du = 3x2 dx, and when x = 0, u = 1 and when x = 1, u = 2. So
dx =
= =
iv) dx = dx
= 2 ln 3 – ln 2 – 2 ln2 + ln 1
= 2 ln 3 – 3 ln 2 (remember that ln 1 = 0)
v) xex dx = – ex dx = e – e + 1 = 1
9.3.12B.
If f(x) is an even function and g(x) is an odd function, show that
i) = 2 ii) = 0
Solution
i) =+
Now put u = – x in the first integral to get
==
since f(x) is an even function. So
=+
= +
= +
= 2
ii) = + = +
= ( +
since g(x) is an odd function
= – + = 0
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