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FIRST YEAR CALCULUS

W W L CHEN

c© W W L Chen, 1994, 2008.

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Chapter 10

TECHNIQUES OF INTEGRATION

10.1. Integration by Substitution

In this section, we discuss how we can use the Chain rule in differentiation to help solve problems inintegration. This technique is usually called integration by substitution. As we shall not prove any resulthere, our discussion will be only heuristic.

We emphasize that the technique does not always work. First of all, we have little or no knowledge ofthe antiderivatives of many functions. Secondly, there is no simple routine that we can describe to helpus find a suitable substitution even in the cases where the technique works. On the other hand, whenthe technique does work, there may well be more than one suitable substitution!

Occasionally, the possibility of substitution may not be immediately obvious, and a certain amount oftrial and error does occur. The fact that one substitution does not appear to work does not mean thatthe method fails. It may very well be the case that we have used a bad substitution.

INTEGRATION BY SUBSTITUTION – VERSION 1. If we make a substitution x = g(u), thendx = g′(u) du, and ∫

f(x) dx =∫f(g(u))g′(u) du.

Example 10.1.1. Consider the indefinite integral∫1√

1− x2dx.

If we make a substitution x = sinu, then dx = cosudu, and∫1√

1− x2dx =

∫cosu√

1− sin2 udu =

∫du = u+ C = sin−1 x+ C.

Chapter 10 : Techniques of Integration page 1 of 26

First Year Calculus c© W W L Chen, 1994, 2008

On the other hand, if we make a substitution x = cos v, then dx = − sin v dv, and∫1√

1− x2dx = −

∫sin v√

1− cos2 vdv = −

∫dv = −v + C = − cos−1 x+ C.

Example 10.1.2. Consider the indefinite integral∫1

1 + x2dx.

If we make a substitution x = tanu, then dx = sec2 udu, and∫1

1 + x2dx =

∫sec2 u

1 + tan2 udu =

∫du = u+ C = tan−1 x+ C.

On the other hand, if we make a substitution x = cot v, then dx = − csc2 v dv, and∫1

1 + x2dx = −

∫csc2 v

1 + cot2 vdv = −

∫dv = −v + C = − cot−1 x+ C.

Example 10.1.3. Consider the indefinite integral∫x√x+ 1 dx.

If we make a substitution x = u2 − 1, then dx = 2udu, and∫x√x+ 1 dx =

∫2(u2 − 1)u2 du = 2

∫u4 du− 2

∫u2 du

=25u5 − 2

3u3 + C =

25

(x+ 1)5/2 − 23

(x+ 1)3/2 + C.

On the other hand, if we make a substitution x = v − 1, then dx = dv, and∫x√x+ 1 dx =

∫(v − 1)v1/2 dv =

∫v3/2 dv −

∫v1/2 dv

=25v5/2 − 2

3v3/2 + C =

25

(x+ 1)5/2 − 23

(x+ 1)3/2 + C.

We can confirm that the indefinite integral is correct by checking that

ddx

(25

(x+ 1)5/2 − 23

(x+ 1)3/2 + C

)= x√x+ 1.

INTEGRATION BY SUBSTITUTION – VERSION 2. Suppose that a function f(x) can bewritten in the form f(x) = g(h(x))h′(x). If we make a substitution u = h(x), then du = h′(x) dx, and∫

f(x) dx =∫g(h(x))h′(x) dx =

∫g(u) du.

Remark. Note that in Version 1, the variable x is initially written as a function of the new variable u,whereas in Version 2, the new variable u is written as a function of x. The difference, however, isminimal, as the substitution x = g(u) in Version 1 has to be invertible to enable us to return from thenew variable u to the original variable x at the end of the process.



Example 10.1.4. Consider the indefinite integral∫x2ex

3dx.

Note first of all that the derivative of the function x3 is equal to 3x2, so it is convenient to make thesubstitution u = x3. Then du = 3x2 dx, and∫

x2ex3

dx =13

∫3x2ex

3dx =

13

∫eu du =

13

eu + C =13

ex3

+ C.

A somewhat more complicated alternative is to note that the derivative of the function ex3

is equal to3x2ex

3, so it is convenient to make the substitution v = ex

3. Then dv = 3x2ex

3dx, and∫

x2ex3

dx =13

∫3x2ex

3dx =

13

∫dv =

13v + C =

13

ex3

+ C.

Example 10.1.5. Consider the indefinite integral∫x(x2 + 3)4 dx.

Note first of all that the derivative of the function x2 + 3 is equal to 2x, so it is convenient to make thesubstitution u = x2 + 3. Then du = 2xdx, and∫

x(x2 + 3)4 dx =12

∫2x(x2 + 3)4 dx =

12

∫u4 du =

110u5 + C =

110

(x2 + 3)5 + C.

Example 10.1.6. Consider the indefinite integral∫1

x log xdx.

Note first of all that the derivative of the function log x is equal to 1/x, so it is convenient to make thesubstitution u = log x. Then du = (1/x) dx, and∫

1x log x

dx =∫

1u

du = log |u|+ C = log | log x|+ C.

Example 10.1.7. Consider the indefinite integral∫tan3 x sec2 x dx.

Note first of all that the derivative of the function tanx is equal to sec2 x, so it is convenient to makethe substitution u = tanx. Then du = sec2 xdx, and∫

tan3 x sec2 x dx =∫u3 du =

14u4 + C =

14

tan4 x+ C.



Example 10.1.8. Consider the indefinite integral∫sin3 x cos3 xdx.

Note first of all that the derivative of the function sinx is equal to cosx, so it is perhaps convenient tomake the substitution u = sinx. Then du = cosxdx, and∫

sin3 x cos3 xdx =∫u3(1− u2) du =

∫(u3 − u5) du =

u4

4− u6

6+ C =

sin4 x

4− sin6 x

6+ C.

Alternatively, note that the derivative of the function cosx is equal to − sinx, so it is convenient to makethe substitution v = cosx. Then dv = − sinxdx, and∫

sin3 x cos3 x dx =∫−(1− v2)v3 dv =

∫(v5 − v3) dv =

v6

6− v4

4+ C ′ =

cos6 x

6− cos4 x

4+ C ′.

It can be checked that

sin4 x

4− sin6 x

6=

cos6 x

6− cos4 x

4+

112.

Example 10.1.9. Recall Example 10.1.1. Since∫1√

1− x2dx = sin−1 x+ C,

we have ∫ 1/2

0

1√1− x2

dx =[

sin−1 x

]1/2

0

= sin−1 12− sin−1 0 =

π

6.

Note that we have in fact used the substitution x = sinu to show that∫1√

1− x2dx =

∫du = u+ C,

followed by an inverse substitution u = sin−1 x. Here, we need to make the extra step of substituting thevalues x = 0 and x = 1/2 to the indefinite integral sin−1 x. Observe, however, that with the substitutionx = sinu, the variable x increases from 0 to 1/2 as the variable u increases from 0 to π/6. But then

∫ π/6

0

du =[u

]π/60

=π

6=∫ 1/2

0

1√1− x2

dx,

so it appears that we do not need the inverse substitution u = sin−1 x. Perhaps we can directly substituteu = 0 and u = π/6 to the indefinite integral u.

DEFINITE INTEGRAL BY SUBSTITUTION – VERSION 1. Suppose that a substitutionx = g(u) satisfies the following conditions:(a) There exist α, β ∈ R such that g(α) = A and g(β) = B.(b) The derivative g′(u) > 0 for every u satisfying α < u < β.Then dx = g′(u) du, and ∫ B

A

f(x) dx =∫ β

α

f(g(u))g′(u) du.



Remark. If condition (b) above is replaced by the condition that the derivative g′(u) < 0 for every usatisfying β < u < α, then the same conclusion holds if we adopt the convention that

∫ β

α

f(g(u))g′(u) du = −∫ α

β

f(g(u))g′(u) du.

Example 10.1.10. To calculate the definite integral∫ 1

0

11 + x2

dx,

we can use the substitution x = tanu, so that dx = sec2 udu. Note that tan 0 = 0 and tan(π/4) = 1,and that sec2 u > 0 whenever 0 < u < π/4. It follows that

∫ 1

0

11 + x2

dx =∫ π/4

0

sec2 u

1 + tan2 udu =

∫ π/4

0

du =[u

]π/40

=π

4− 0 =

π

4.

We can compare this to first observing Example 10.1.2, so that

∫ 1

0

11 + x2

dx =[

tan−1 x

]1

0

= tan−1 1− tan−1 0 =π

4− 0 =

π

4.


0

x√x+ 1 dx,

we can use the substitution x = g(u) = u2 − 1, so that dx = 2udu. Note that g(1) = 0 and g(2) = 3,and that g′(u) = 2u > 0 whenever 1 < u < 2. It follows that

∫ 3

0

x√x+ 1 dx =

∫ 2

1

2(u2 − 1)u2 du =[

25u5 − 2

3u3

]2

1

=(

645− 16

3

)−(

25− 2

3

)=

625− 14

3=

11615

.

DEFINITE INTEGRAL BY SUBSTITUTION – VERSION 2. Suppose that a substitutionu = h(x) satisfies the following conditions:(a) There exists a function g(u) such that f(x) = g(h(x))h′(x) for every x ∈ [A,B].(b) The derivative h′(x) > 0 for every x satisfying A < x < B.Then du = h′(x) dx, and

∫ B

A

f(x) dx =∫ B

A

g(h(x))h′(x) dx =∫ h(B)

h(A)

g(u) du.

Remark. If condition (b) above is replaced by the condition that the derivative h′(x) < 0 for every xsatisfying A < x < B, then the same conclusion holds if we adopt the convention that

∫ h(B)

h(A)

g(u) du = −∫ h(A)

h(B)

g(u) du.




0

x(x2 + 3)4 dx,

we can use the substitution u = h(x) = x2 + 3, so that du = 2xdx. Note that h(0) = 3 and h(1) = 4,and that h′(x) = 2x > 0 whenever 0 < x < 1. It follows that∫ 1

0

x(x2 + 3)4 dx =12

∫ 4

3

u4 dx =12

[u5

5

]4

3

=12

(1024

5− 243

5

)=

78110

.

We can compare this to first observing Example 10.1.4, so that∫ 1

0

x(x2 + 3)4 dx =[

110

(x2 + 3)5

]1

0

=102410− 243

10=

78110

.


2

1x log x

dx,

we can use the substitution u = h(x) = log x, so that du = h′(x) dx, where h′(x) = 1/x > 0 whenever2 < x < 4. Note also that h(2) = log 2 and h(4) = log 4. It follows that∫ 4

2

1x log x

dx =∫ log 4

log 2

1u

du =[

log |u|]log 4

log 2

= log log 4− log log 2 = log(

log 4log 2

)= log 2.

10.2. Integration by Parts

Recall the Product rule for differentiation, that

(uv)′ = uv′ + vu′.

Integrating with respect to x, we obtain∫(uv)′ dx =

∫uv′ dx+

∫vu′ dx.

Now the indefinite integral on the left hand side is of the form uv. Rewriting this equation, we have∫uv′ dx = uv −

∫vu′ dx. (1)

Equation (1) is called the formula for integration by parts for indefinite integrals. It is very useful if theindefinite integral

∫vu′ dx is much easier to calculate than the indefinite integral

∫uv′ dx.

Example 10.2.1. Consider the indefinite integral∫xex dx.

Writing u = x and v′ = ex, we have ∫uv′ dx =

∫xex dx.



Furthermore, v = ex and u′ = 1. It follows that

uv −∫vu′ dx = xex −

∫ex dx = xex − ex + C.

Hence ∫xex dx = xex − ex + C.

Example 10.2.2. Consider the indefinite integral∫log xdx.

Writing u = log x and v′ = 1, we have ∫uv′ dx =

∫log xdx.

Furthermore,

v = x and u′ =1x.

It follows that

uv −∫vu′ dx = x log x−

∫x

1x

dx = x log x− x+ C.

Hence ∫log x dx = x log x− x+ C.

Example 10.2.3. Consider the indefinite integral∫ex sinxdx.

Writing u = ex and v′ = sinx, we have∫uv′ dx =

∫ex sinxdx.

Furthermore, v = − cosx and u′ = ex. It follows that

uv −∫vu′ dx = −ex cosx+

∫ex cosxdx.

Hence ∫ex sinxdx = −ex cosx+

∫ex cosx dx. (2)

We now need to study the indefinite integral∫ex cosxdx.



Writing u = ex and v′ = cosx, we have∫uv′ dx =

∫ex cosx dx.

Furthermore, v = sinx and u′ = ex. It follows that

uv −∫vu′ dx = ex sinx−

∫ex sinx dx.

Hence ∫ex cosxdx = ex sinx−

∫ex sinx dx. (3)

It looks like we are back to the same old problem. However, if we combine (2) and (3), then we obtain∫ex sinxdx = −ex cosx+ ex sinx−

∫ex sinx dx,

so that

2∫

ex sinxdx = ex sinx− ex cosx = ex(sinx− cosx).

Adding an arbitrary constant, which we may in view of Proposition 9C, we have∫ex sinx dx =

12

ex(sinx− cosx) + C.

Example 10.2.4. Consider the indefinite integral∫x3 cosxdx.

Writing u = x3 and v′ = cosx, we have∫uv′ dx =

∫x3 cosxdx.

Furthermore, v = sinx and u′ = 3x2. It follows that

uv −∫vu′ dx = x3 sinx− 3

∫x2 sinxdx.

Hence ∫x3 cosxdx = x3 sinx− 3

∫x2 sinxdx. (4)

We now need to study the indefinite integral∫x2 sinx dx.

Writing u = x2 and v′ = sinx, we have∫uv′ dx =

∫x2 sinx dx.



Furthermore, v = − cosx and u′ = 2x. It follows that

uv −∫vu′ dx = −x2 cosx+ 2

∫x cosx dx.

Hence ∫x2 sinxdx = −x2 cosx+ 2

∫x cosxdx. (5)

Combining (4) and (5), we have∫x3 cosxdx = x3 sinx+ 3x2 cosx− 6

∫x cosxdx. (6)

We now need to study the indefinite integral ∫x cosxdx.

Writing u = x and v′ = cosx, we have ∫uv′ dx =

∫x cosx dx.

Furthermore, v = sinx and u′ = 1. It follows that

uv −∫vu′ dx = x sinx−

∫sinx dx.

Hence ∫x cosxdx = x sinx−

∫sinx dx. (7)

Combining (6) and (7), we have∫x3 cosxdx = x3 sinx+ 3x2 cosx− 6x sinx+ 6

∫sinxdx

= x3 sinx+ 3x2 cosx− 6x sinx− 6 cosx+ C.

The technique is also valid for definite integrals, in view of the first Fundamental theorem of integralcalculus. For definite integrals over the interval [A,B], we have∫ B

A

uv′ dx =[uv

]x=B

x=A

−∫ B

A

vu′ dx. (8)

Equation (8) is called the formula for integration by parts for definite integrals.

Example 10.2.5. Consider the definite integral∫ π/2

0

x3 cosxdx.

Writing u = x3 and v′ = cosx, we have∫ π/2

0

uv′ dx =∫ π/2

0

x3 cosx dx.



Furthermore, v = sinx and u′ = 3x2. It follows that[uv

]x=π/2

x=0

−∫ π/2

0

vu′ dx =[x3 sinx

]π/20

− 3∫ π/2

0

x2 sinx dx.

Hence ∫ π/2

0

x3 cosx dx =[x3 sinx

]π/20

− 3∫ π/2

0

x2 sinxdx =π3

8− 3

∫ π/2

0

x2 sinxdx. (9)

We now need to study the definite integral ∫ π/2

0

x2 sinxdx.

Writing u = x2 and v′ = sinx, we have∫ π/2

0

uv′ dx =∫ π/2

0

x2 sinxdx.

Furthermore, v = − cosx and u′ = 2x. It follows that[uv

]x=π/2

x=0

−∫ π/2

0

vu′ dx =[−x2 cosx

]π/20

+ 2∫ π/2

0

x cosx dx.

Hence ∫ π/2

0

x2 sinxdx =[−x2 cosx

]π/20

+ 2∫ π/2

0

x cosxdx = 2∫ π/2

0

x cosxdx. (10)

Combining (9) and (10), we have∫ π/2

0

x3 cosxdx =π3

8− 6

∫ π/2

0

x cosx dx. (11)

We now need to study the definite integral ∫ π/2

0

x cosx dx.

Writing u = x and v′ = cosx, we have∫ π/2

0

uv′ dx =∫ π/2

0

x cosxdx.

Furthermore, v = sinx and u′ = 1. It follows that[uv

]x=π/2

x=0

−∫ π/2

0

vu′ dx =[x sinx

]π/20

−∫ π/2

0

sinxdx.

Hence ∫ π/2

0

x cosx dx =[x sinx

]π/20

−∫ π/2

0

sinxdx =π

2−∫ π/2

0

sinx dx. (12)



Combining (11) and (12), we have∫ π/2

0

x3 cosxdx =π3

8− 3π + 6

∫ π/2

0

sinxdx =π3

8− 3π + 6

[− cosx

]π/20

=π3

8− 3π + 6.

10.3. Trigonometric Integrals

In this section, we consider integrals involving the six trigonometric functions sinx, cosx, tanx, cotx,secx and cscx. If we consider differentiation formulas involving these functions, then we can dividethese into three groups: (a) sinx and cosx; (b) tanx and secx; and (c) cotx and cscx. Note that thederivative of any of these functions can be expressed in terms of the two functions in the group to whichit belongs. This division is also substantiated by integral formulas.

It follows that given any indefinite integral ∫f(x) dx,

where the integrand f(x) involves trigonometric functions, it may be beneficial to try first to expressf(x) in terms of trigonometric functions from only one of these three groups.

Example 10.3.1. Consider the indefinite integral∫tanx+ sec3 x cotx

cos2 xdx =

∫ (tanxcos2 x

+sec3 x cotx

cos2 x

)dx

=∫ (

sinxcos3 x

+sec5 x

tanx

)dx =

∫sinx

cos3 xdx+

∫sec5 x

tanxdx.

Note that we can also write∫sinx

cos3 xdx =

∫tanx sec2 xdx =

12

tan2 x+ C.

However, the indefinite integral ∫sec5 x

tanxdx

does not appear to be so simple.

Let us consider first integrals involving sinx and cosx. Consider an integral of the form∫sinm x cosn xdx.

When m = 1, the integral is simple to evaluate. Clearly∫sinx cosn x dx = − 1

n+ 1cosn+1 x+ C if n 6= −1,

and ∫sinx cos−1 x dx = − log | cosx|+ C.



When n = 1, the integral is also simple to evaluate. Clearly∫sinm x cosxdx =

1m+ 1

sinm+1 x+ C if m 6= −1,

and ∫sin−1 x cosx dx = log | sinx|+ C.

In the general case, we may use standard trigonometric formulas like

sin2 x+ cos2 x = 1, (13)

sin 2x = 2 sinx cosx, (14)

cos 2x = cos2 x− sin2 x. (15)

Note also that combining (13) and (15), we have

cos 2x = 2 cos2 x− 1 = 1− 2 sin2 x. (16)

Example 10.3.2. Consider the indefinite integral∫sin5 x dx.

Using (13), we can write

sin5 x = sin4 x sinx = (1− cos2 x)2 sinx = (1− 2 cos2 x+ cos4 x) sinx,

so that ∫sin5 xdx =

∫(1− 2 cos2 x+ cos4 x) sinxdx

=∫

sinxdx− 2∫

cos2 x sinx dx+∫

cos4 x sinx dx

= − cosx+23

cos3 x− 15

cos5 x+ C.



sin3 x cos3 x = cos2 x sin3 x cosx = (1− sin2 x) sin3 x cosx = sin3 x cosx− sin5 x cosx,

so that ∫sin3 x cos3 xdx =

∫(sin3 x cosx− sin5 x cosx) dx

=∫

sin3 x cosxdx−∫

sin5 x cosx dx

=14

sin4 x− 16

sin6 x+ C.



Example 10.3.4. Consider the indefinite integral∫sin4 4xdx.


sin4 4x =14

(1− cos 8x)2 =14

(1− 2 cos 8x+ cos2 8x)

=14

(1− 2 cos 8x+

12

(1 + cos 16x))

=38− 1

2cos 8x+

18

cos 16x,

so that ∫sin4 4xdx =

∫ (38− 1

2cos 8x+

18

cos 16x)

dx

=38

∫dx− 1

2

∫cos 8xdx+

18

∫cos 16x dx

=38x− 1

16sin 8x+

1128

sin 16x+ C.


Using (14) and (16), we can write

sin2 x cos4 x = cos2 x(sinx cosx)2 =18

(1 + cos 2x) sin2 2x =18

sin2 2x+18

cos 2x sin2 2x

=116

(1− cos 4x) +18

cos 2x sin2 2x =116− 1

16cos 4x+

18

cos 2x sin2 2x,

so that ∫sin2 x cos4 x dx =

∫ (116− 1

16cos 4x+

18

cos 2x sin2 2x)

dx

=116

∫dx− 1

16

∫cos 4x dx+

18

∫cos 2x sin2 2x dx

=116x− 1

64sin 4x+

148

sin3 2x+ C.

Let us consider next integrals involving tanx and secx. Consider an integral of the form∫tanm x secn xdx.

When m = 1, the integral is simple to evaluate. Clearly∫tanx secn xdx =

1n

secn x+ C if n 6= 0,

and ∫tanxdx = − log | cosx|+ C.



When n = 2, the integral is also simple to evaluate. Clearly∫tanm x sec2 xdx =

1m+ 1

tanm+1 x+ C if m 6= −1,

and ∫tan−1 x sec2 x dx = log | tanx|+ C.

In the general case, we may use standard trigonometric formulas like

1 + tan2 x = sec2 x. (17)

Example 10.3.6. Consider the indefinite integral∫tan3 x dx.


tan3 x = tan2 x tanx = (sec2 x− 1) tanx = sec2 x tanx− tanx,

so that ∫tan3 x dx =

∫(sec2 x tanx− tanx) dx

=∫

sec2 x tanx dx−∫

tanxdx

=12

tan2 x+ log | cosx|+ C.

Example 10.3.7. Consider the indefinite integral∫tan4 x dx.


tan4 x = tan2 x tan2 x = (sec2 x− 1) tan2 x = sec2 x tan2 x− tan2 x = sec2 x tan2 x− sec2 x+ 1,

so that ∫tan4 x dx =

∫(sec2 x tan2 x− sec2 x+ 1) dx

=∫

sec2 x tan2 xdx−∫

sec2 x dx+∫

dx

=13

tan3 x− tanx+ x+ C.



Example 10.3.8. Consider the indefinite integral∫sec3 xdx.

Writing u = secx and v′ = sec2 x, we have∫uv′ dx =

∫sec3 xdx.

Furthermore, v = tanx and u′ = tanx secx. It follows that

uv −∫vu′ dx = secx tanx−

∫tan2 x secx dx.

Hence ∫sec3 xdx = secx tanx−

∫tan2 x secx dx. (18)

We now need to study the indefinite integral∫tan2 x secx dx.


tan2 x secx = (sec2 x− 1) secx = sec3 x− secx,

so that ∫tan2 x secx dx =

∫(sec3 x− secx) dx =

∫sec3 x dx−

∫secxdx. (19)

Combining (18) and (19), we have∫sec3 xdx = secx tanx−

∫sec3 xdx+

∫secxdx,

so that ∫sec3 xdx =

12

secx tanx+12

∫secx dx =

12

secx tanx+12

log | secx+ tanx|+ C.

Example 10.3.9. Consider the indefinite integral∫tan2 x sec3 xdx.

Writing u = tan2 secx and v′ = sec2 x, we have∫uv′ dx =

∫tan2 x sec3 x dx.

Furthermore, v = tanx and u′ = 2 tanx sec3 x+ tan3 x secx. It follows that

uv −∫vu′ dx = tan3 x secx−

∫(2 tan2 x sec3 x+ tan4 x secx) dx.



Hence ∫tan2 x sec3 xdx = tan3 x secx−

∫(2 tan2 x sec3 x+ tan4 x secx) dx

= tan3 x secx− 2∫

tan2 x sec3 xdx−∫

tan4 x secx dx. (20)

We now need to study the indefinite integral∫tan4 x secx dx.

Using (17), we can write (the reader should check this)

tan4 x secx = tan2 x sec3 x− sec3 x+ secx,

so that ∫tan4 x secxdx =

∫tan2 x sec3 xdx−

∫sec3 xdx+

∫secxdx. (21)

Combining (20) and (21), we have∫tan2 x sec3 x dx =

14

tan3 x secx+14

∫sec3 xdx− 1

4

∫secxdx

=14

tan3 x secx+18

tanx secx− 18

log | tanx+ secx|+ C.

Occasionally, it may be necessary to convert an expression involving tanx and secx to one involvingsinx and cosx instead.

Example 10.3.10. Consider the indefinite integral∫tan4 7xsec5 7x

dx.

Here the identity (17) does not help very much. However, we have

tan4 7xsec5 7x

= sin4 7x cos 7x,

so that ∫tan4 7xsec5 7x

dx =∫

sin4 7x cos 7xdx =135

sin5 7x+ C.

Let us consider finally integrals involving cotx and cscx. Consider an integral of the form∫cotm x cscn xdx.

When m = 1, the integral is simple to evaluate. Clearly∫cotx cscn xdx = − 1

ncscn x+ C if n 6= 0,



and ∫cotx dx = log | sinx|+ C.

When n = 2, the integral is also simple to evaluate. Clearly∫cotm x csc2 x dx = − 1

m+ 1cotm+1 x+ C if m 6= −1,

and ∫cot−1 x csc2 xdx = − log | cotx|+ C.

The details are similar to the case of tanx and secx.

10.4. Trigonometric Substitutions

In this section, we shall consider techniques to handle integrals involving square roots of the form√a2 − b2x2,

√a2 + b2x2 or

√b2x2 − a2. Without loss of generality, assume that a, b > 0.

Let us consider first the case√a2 − b2x2. If we use the substitution

x =a

bsin θ,

then √a2 − b2x2 =

√a2(1− sin2 θ) =

√a2 cos2 θ = a| cos θ|,

while

dx =a

bcos θ dθ.


4− x2dx.

If we use the substitution x = 2 sin θ, then√4− x2 = 2| cos θ| and dx = 2 cos θ dθ,

so that ∫1√

4− x2dx =

∫1

2| cos θ|2 cos θ dθ.

Suppose that cos θ > 0. Then∫1√

4− x2dx =

∫dθ = θ + C = sin−1

(x2

)+ C.



Example 10.4.2. Consider the indefinite integral∫x3√

9− 4x2 dx.

If we use the substitution x = 32 sin θ, then√

9− 4x2 = 3| cos θ| and dx =32

cos θ dθ.

Suppose that cos θ > 0. Then∫x3√

9− 4x2 dx =24316

∫sin3 θ cos2 θ dθ

=24316

∫(1− cos2 θ) sin θ cos2 θ dθ

=24316

∫sin θ cos2 θ dθ − 243

16

∫sin θ cos4 θ dθ

= −8116

cos3 θ +24380

cos5 θ + C.

Next, note that cos2 θ = 1− sin2 θ = 1− 49x

2, so that∫x3√

9− 4x2 dx = −8116

(1− 4

9x2

)3/2

+24380

(1− 4

9x2

)5/2

+ C.

Let us consider next the case√a2 + b2x2. If we use the substitution

x =a

btan θ,

then √a2 + b2x2 =

√a2(1 + tan2 θ) =

√a2 sec2 θ = a| sec θ|,

while

dx =a

bsec2 θ dθ.

Example 10.4.3. Consider the indefinite integral∫x2√

1 + x2 dx.

If we use the substitution x = tan θ, then√1 + x2 = | sec θ| and dx = sec2 θ dθ.

Suppose that sec θ > 0. Then ∫x2√

1 + x2 dx =∫

tan2 θ sec3 θ dθ.

We have shown earlier that∫tan2 θ sec3 θ dθ =

14

tan3 θ sec θ +18

tan θ sec θ − 18

log | tan θ + sec θ|+ C.



Next, note that sec2 θ = 1 + tan2 θ = 1 + x2, so that∫x2√

1 + x2 dx =14x3(1 + x2)1/2 +

18x(1 + x2)1/2 − 1

8log |x+ (1 + x2)1/2|+ C.

Let us consider finally the case√b2x2 − a2. If we use the substitution

x =a

bsec θ,

then √b2x2 − a2 =

√a2(sec2 θ − 1) =

√a2 tan2 θ = a| tan θ|,

while

dx =a

btan θ sec θ dθ.

Example 10.4.4. Consider the indefinite integral∫ √x2 − 4x

dx.

If we use the substitution x = 2 sec θ, then√x2 − 4 = 2| tan θ| and dx = 2 tan θ sec θ dθ.

Suppose that tan θ > 0. Then∫ √x2 − 4x

dx = 2∫

tan2 θ dθ = 2∫

(sec2 θ − 1) dθ = 2∫

sec2 θ dθ − 2∫

dθ = 2 tan θ − 2θ + C.

Next, note that

tan2 θ = sec2 θ − 1 =14x2 − 1 and θ = sec−1

(x2

),

so that ∫ √x2 − 4x

dx = 2(

14x2 − 1

)1/2

− 2 sec−1(x

2

)+ C =

√x2 − 4− 2 sec−1

(x2

)+ C.

10.5. Completing Squares

In this section, we shall consider techniques to handle integrals involving square roots of the form√αx2 + βx+ γ, where α 6= 0. Our task is to show that such integrals can be reduced to integrals

discussed in the previous section.

Note that

αx2 + βx+ γ = α

(x2 +

β

αx+

γ

α

)= α

(x2 +

β

αx+

β2

4α2

)+(γ − b2

4α

)= α

(x+

β

2α

)2

+(γ − β2

4α

).



Suppose first of all that we use a substitution

y = x+β

2α.

Then dy = dx and

αx2 + βx+ γ = αy2 + δ,

where

δ = γ − β2

4α.

It now follows that√αx2 + βx+ γ is of the form

√a2 − b2y2 if α < 0 and δ > 0,√a2 + b2y2 if α > 0 and δ > 0,√b2y2 − a2 if α > 0 and δ < 0.


3− 2x− x2dx.

We have

3− 2x− x2 = −(x2 + 2x− 3) = −(x2 + 2x+ 1) + 4 = −(x+ 1)2 + 4 = −y2 + 4,

where we use the substitution y = x+ 1. Note that α = −1 < 0 and δ = 4 > 0. Then∫1√

3− 2x− x2dx =

∫1√

4− y2dx.

We have shown earlier that ∫1√

4− y2dy = sin−1

(y2

)+ C.

It follows that ∫1√

3− 2x− x2dx = sin−1

(x+ 1

2

)+ C.

Example 10.5.2. Consider the indefinite integral∫ √x2 − 4xx− 2

dx.

We have

x2 − 4x = (x2 − 4x+ 4)− 4 = (x− 2)2 − 4 = y2 − 4,

where we use the substitution y = x− 2. Note that α = 1 > 0 and δ = −4 < 0. Then∫ √x2 − 4xx− 2

dx =∫ √

y2 − 4y

dy.



We have shown earlier that ∫ √y2 − 4y

dy =√y2 − 4− 2 sec−1

(y2

)+ C.

It follows that∫ √x2 − 4xx− 2

dx =√

(x− 2)2 − 4− 2 sec−1

(x− 2

2

)+ C =

√x2 − 4x− 2 sec−1

(x− 2

2

)+ C.

10.6. Partial Fractions

In this section, we shall consider techniques to handle integrals of the form∫p(x)q(x)

dx,

where p(x) and q(x) are polynomials in x.

If the degree of p(x) is not smaller than the degree of q(x), then we can always find polynomials a(x)and r(x) such that

p(x)q(x)

= a(x) +r(x)q(x)

,

where r(x) = 0 or has degree smaller than the degree of q(x).

Example 10.6.1. Consider the indefinite integral∫x5 + 2x4 + 4x3 + x+ 1

x2 + x+ 1dx.

Note that

x5 + 2x4 + 4x3 + x+ 1x2 + x+ 1

= (x3 + x2 + 2x− 3) +2x+ 4

x2 + x+ 1,

so that ∫x5 + 2x4 + 4x3 + x+ 1

x2 + x+ 1dx =

∫(x3 + x2 + 2x− 3) dx+

∫2x+ 4

x2 + x+ 1dx.

It does not take a genius to work out the indefinite integral∫(x3 + x2 + 2x− 3) dx.

We can therefore restrict our attention to the case when the polynomial p(x) is of lower degree thanthe polynomial q(x).

The first step is to factorize the polynomial q(x) into a product of irreducible factors. It is a funda-mental result in algebra that a real polynomial q(x) can be factorized into a product of irreducible linearfactors and quadratic factors with real coefficients.



Example 10.6.2. Suppose that q(x) = x4 − 4x3 + 5x2 − 4x + 4. Then q(x) can be factorized into aproduct of irreducible linear factors in the form (x− 2)2(x2 + 1).

Suppose that a linear factor (ax+ b) occurs n times in the factorization of q(x). Then we write downa decomposition

A1

ax+ b+

A2

(ax+ b)2+ . . .+

An(ax+ b)n

,

where the constants A1, . . . , An will be determined later. Suppose that a quadratic factor (ax2 + bx+ c)occurs n times in the factorization of q(x). Then we write down a decomposition

A1x+B1

ax2 + bx+ c+

A2x+B2

(ax2 + bx+ c)2+ . . .+

Anx+Bn(ax2 + bx+ c)n

,

where the constants A1, . . . , An and B1, . . . , Bn will be determined later. We proceed to add all thedecompositions and equate their sum to

p(x)q(x)

,

and then calculate all the constants by equating coefficients.

Example 10.6.3. Suppose that

p(x)q(x)

=2x3 − 11x2 + 17x− 16x4 − 4x3 + 5x2 − 4x+ 4

=2x3 − 11x2 + 17x− 16

(x− 2)2(x2 + 1).

We now write

2x3 − 11x2 + 17x− 16(x− 2)2(x2 + 1)

=c1

x− 2+

c2(x− 2)2

+c3x+ c4x2 + 1

.

Now

c1x− 2

+c2

(x− 2)2+c3x+ c4x2 + 1

=c1(x− 2)(x2 + 1) + c2(x2 + 1) + (c3x+ c4)(x− 2)2

(x− 2)2(x2 + 1),

so that

c1(x− 2)(x2 + 1) + c2(x2 + 1) + (c3x+ c4)(x− 2)2 = 2x3 − 11x2 + 17x− 16.

Note now that

c1(x− 2)(x2 + 1) + c2(x2 + 1) + (c3x+ c4)(x− 2)2

= c1(x3 − 2x2 + x− 2) + c2(x2 + 1) + c3(x3 − 4x2 + 4x) + c4(x2 − 4x+ 4)= (c1 + c3)x3 + (−2c1 + c2 − 4c3 + c4)x2 + (c1 + 4c3 − 4c4)x+ (−2c1 + c2 + 4c4).

Equating coefficients, we have

c1 + c3 = 2,−2c1 + c2 − 4c3 + c4 = −11,c1 + 4c3 − 4c4 = 17,

−2c1 + c2 + 4c4 = −16.



This system has solution c1 = 1, c2 = −2, c3 = 1 and c4 = −3. Hence

2x3 − 11x2 + 17x− 16x4 − 4x3 + 5x2 − 4x+ 4

=1

x− 2− 2

(x− 2)2+

x− 3x2 + 1

,

so that ∫2x3 − 11x2 + 17x− 16x4 − 4x3 + 5x2 − 4x+ 4

dx =∫

1x− 2

dx−∫

2(x− 2)2

dx+∫

x− 3x2 + 1

dx.

We shall calculate the three indefinite integrals on the right hand side later.

To calculate the indefinite integrals that arise, note that these indefinite integrals are of the form∫A

(ax+ b)kdx, (22)

or ∫Ax+B

(ax2 + bx+ c)kdx, (23)

where A and B are constants and k is a positive integer. The integral (22) is simple. If k 6= 1, then wehave ∫

A

(ax+ b)kdx = − A

(k − 1)a(ax+ b)k−1+ C.

On the other hand, we have ∫A

ax+ bdx =

A

alog |ax+ b|+ C.

The integral (23) is a bit more complicated. Note that∫Ax+B

(ax2 + bx+ c)kdx =

A

2a

∫2ax+ b

(ax2 + bx+ c)kdx+

(B − Ab

2a

)∫1

(ax2 + bx+ c)kdx.

The first integral on the right hand side is simple. If k 6= 1, then we have∫2ax+ b

(ax2 + bx+ c)kdx = − 1

(k − 1)(ax2 + bx+ c)k−1+ C.

On the other hand, we have ∫2ax+ b

ax2 + bx+ cdx = log |ax2 + bx+ c|+ C.

It remains to study the integral ∫1

(ax2 + bx+ c)kdx.

To do this, we may try the technique of completing squares as described in the previous section, andthen use a trigonometric substitution to find the integral.



Example 10.6.4. Let us continue the discussion of Example 10.6.3. Note that∫1

x− 2dx = log |x− 2|+ C and

∫1

(x− 2)2dx = − 1

x− 2+ C.

On the other hand, we have∫x− 3x2 + 1

dx =12

∫2x

x2 + 1dx− 3

∫1

x2 + 1dx.

Clearly ∫2x

x2 + 1dx = log |x2 + 1|+ C.

Using the substitution x = tan θ, we have∫1

x2 + 1dx =

∫sec2 θ

1 + tan2 θdθ =

∫dθ = θ + C = tan−1 x+ C.

It follows that∫2x3 − 11x2 + 17x− 16x4 − 4x3 + 5x2 − 4x+ 4

dx =∫

1x− 2

dx−∫

2(x− 2)2

dx+∫

x− 3x2 + 1

dx

= log |x− 2|+ 2x− 2

+12

log |x2 + 1| − 3 tan−1 x+ C.

Example 10.6.5. Consider the indefinite integral∫x2 + x− 3

x3 − 2x2 − x+ 2dx.

Note first of all that

x3 − 2x2 − x+ 2 = (x− 2)(x+ 1)(x− 1),

so we consider partial fractions of the form

x2 + x− 3(x− 2)(x+ 1)(x− 1)

=c1

x− 2+

c2x+ 1

+c3

x− 1

=c1(x+ 1)(x− 1) + c2(x− 2)(x− 1) + c3(x− 2)(x+ 1)

(x− 2)(x+ 1)(x− 1).

It follows that

c1(x+ 1)(x− 1) + c2(x− 2)(x− 1) + c3(x− 2)(x+ 1) = x2 + x− 3. (24)

We may equate coefficients and solve for c1, c2, c3. Alternatively, substituting x = 2,−1, 1 into equation(24), we get respectively 3c1 = 3, 6c2 = −3 and −2c3 = −1, so that c1 = 1, c2 = −1/2 and c3 = 1/2.Hence ∫

x2 + x− 3x3 − 2x2 − x+ 2

dx =∫

1x− 2

dx− 12

∫1

x+ 1dx+

12

∫1

x− 1dx

= log |x− 2| − 12

log |x+ 1|+ 12

log |x− 1|+ C.



Example 10.6.6. Consider the indefinite integral∫x6 − 2x4 + x2

dx.

Note that

x6 − 2x4 + x2

= x2 − 1 +x2 − 2x4 + x2

,

so that ∫x6 − 2x4 + x2

dx =∫

(x2 − 1) dx+∫

x2 − 2x4 + x2

dx =13x3 − x+

∫x2 − 2x4 + x2

dx. (25)

Next, we study the integral ∫x2 − 2x4 + x2

dx.

Note first of all that

x4 + x2 = x2(x2 + 1),

so we consider partial fractions of the form

x2 − 2x2(x2 + 1)

=c1x

+c2x2

+c3x+ c4x2 + 1

=c1x(x2 + 1) + c2(x2 + 1) + (c3x+ c4)x2

x2(x2 + 1).

It follows that

c1x(x2 + 1) + c2(x2 + 1) + (c3x+ c4)x2 = x2 − 2.

Equating coefficients, we have

c1 + c3 = 0,c2 + c4 = 1,

c1 = 0,c2 = −2.

This system has solution c1 = 0, c2 = −2, c3 = 0 and c4 = 3. Hence∫x2 − 2x4 + x2

dx = −2∫

1x2

dx+ 3∫

1x2 + 1

dx =2x

+ 3 tan−1 x+ C. (26)

Combining (25) and (26), we obtain∫x6 − 2x4 + x2

dx =13x3 − x+

2x

+ 3 tan−1 x+ C.



Problems for Chapter 10

1. Evaluate each of the following indefinite integrals:

a)∫

sinx cos 7x dx b)∫

e2x cos 3xdx c)∫x2 log x dx

d)∫

cos 2x1− sin 2x

dx e)∫

1√16− 3x+ x2

dx f)∫x sec2 x dx

g)∫

1x2 + 4x− 4

dx h)∫

x2

x3 + 3x2 + 3x+ 1dx i)

∫ √cotx csc4 xdx

j)∫

x2 + 3x− 1x4 + x3 + x2 + x

dx k)∫

log(x6) dx l)∫

sin2 3xdx

m)∫

1x2 − 5x+ 4

dx n)∫

e2x cosxdx o)∫

(x3 +√x) dx

p)∫x4 + x

√x+ 1

xdx q)

∫x2√x− 1 dx r)

∫x√x2 + 4 dx

s)∫

e4x+2 dx t)∫xex

2dx u)

∫log xx

dx

v)∫

(log x)5

xdx w)

∫2x+ 3

x2 + 3x− 4dx x)

∫sin−1 x√

1− x2dx

y)∫

1√a2 − x2

dx z)∫

(√x+ 1)10

√x

dx aa)∫

1x2 + a2

dx

bb)∫x5ex dx cc)

∫1

x2 − 4x+ 3dx dd)

∫xex dx

ee)∫

x− 4(x2 + 4)(x+ 1)

dx

2. Evaluate each of the following definite integrals:

a)∫ 3

2

x(1 + 2x2)4 dx b)∫ 4

1

(√x+ 1√x

)dx c)

∫ 2

1

x2 + 1(x+ 1)4

dx

d)∫ π/4

0

cosx(1 + sinx)2

dx e)∫ 1

0

x√x2 + 1 dx f)

∫ 4

1

e√x

√x

dx

g)∫ π/4

0

cos2 2x dx h)∫ π/2

0

√2− 2 cosx dx i)

∫ π/4

0

x cos 2x dx

j)∫ 1/2

0

x√1− x2

dx


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