AMM 104: Calculus II
Instruction Hours: 45
Pre-requisites: AMM 103
Purpose of the course
To introduce the student to basic integration methods and their applications
Expected Learning Outcomes
By the end of the course the learner should be able to:
i) State the fundamental theorem of integral calculus
ii) Use various integration methods to integrate functions
iii) Apply integration to find the length of an arc, area under a curve and the volume of solid of
revolution
iv) Evaluate definite integrals by trapezoidal and Simpson’s rules
Course Content
Anti-derivative, Fundamental theorem of integration. Techniques of integration: Rules of
integration, Methods of integration: Algebraic substitution, powers of trigonometric functions,
Use of factor and t-formulae, use of inverse trigonometric and hyperbolic functions, Integration
by partial fractions. Integration by parts, Reduction formulae.
Applications of Integration; arc length, area and volume of solids of revolution, mean value and
root-mean square value.
Multiple Integrals: Double integrals, change of the order of integration and the Fubini’s theorem.
Polar and curvilinear coordinates. Moment of inertia. Change of the coordinate system. Triple
integrals, space coordinate systems.
Power series: Maclaurin’s and Taylor’s theorems and their applications to singular integrals.
Teaching / Learning Methodologies: Lectures; Tutorials; Class discussion
Instructional Materials and Equipment: Handouts; White board
Course Assessment: Examination - 70%; Continuous Assessments (Exercises and Tests) - 30%;
Total - 100%
1
Recommended Text Books
1. Tom M. Apostol (2007); Calculus, Volume I, 2nd Ed; Wiley
2. Alex (2009); Calculus Ideas and Applications; John Wiley and Sons
3. Thomas G.B., and Finney R.L (2009) 11th Edition; Calculus and Analytical geometry,
Wesley
4. Hunt Richard A (2009); Calculus; (2nd Edition); Harper Collins College Publishers
Text Books for further Reading
1. Bradley Smith (2009); Calculus; Prentice Hall- Gale.
2. Stein Sheran K. (2008); Calculus with analytic Geometry; McGraw Hill.
3. Boyce William E and DiPrima, Richard C.(2008); Calculus; John Wiley.
4. Hirst K. E and Hirst K. E (2005); Calculus of One Variable; Springer
Integral Calculus
Integration is the process by which we determine functions from their derivatives. Integration
and differentiation are intimately connected. They are reverse processes of each other.
The process of determining a function from its derivative 𝑓(𝑥) and one of its known values has
two steps. The first is to find a formula that gives us all the functions that could possibly have
𝑓(𝑥) as a derivative. These functions are the so – called antiderivatives of 𝑓, and the formula that
gives them all is called the indefinite integral of 𝑓. The second step is to use the known function
value to select the particular antiderivative needed from the indefinite integral.
Finding Antiderivaties – Indefinite Integrals
Definitions
A function 𝐹(𝑥) is an antiderivative of a function 𝑓(𝑥) if
𝐹 ′ (𝑥) = 𝑓(𝑥)
for all 𝑥 in the domain of 𝑓. The set of all antiderivatives of 𝑓 is the indefinite integral of 𝑓 with
respect to 𝑥 denoted by
2
∫ 𝑓(𝑥)𝑑𝑥.
The symbol ∫ is an integral sign. The function 𝑓 is the integrand of the integral and 𝑥 is the
variable of integration. In integral notation we write
∫ 𝑓(𝑥) 𝑑𝑥 = 𝐹(𝑥) + 𝐶.
(1)
The constant C is the constant of integration or the arbitrary constant. Equation (1) is read,
“The indefinite integral of 𝑓 with respect to x is 𝐹(𝑥) + 𝐶". When we find 𝐹(𝑥) + 𝐶, we say that
we have integrated 𝑓 and evaluated the integral.
Example
Evaluate∫ 2𝑥 𝑑𝑥.
Solution
an antiderivative of 2x
∫ 2𝑥 𝑑𝑥 = 𝑥 2 + 𝐶
The formula 𝑥 2 + 𝐶 generates all the antiderivatives of the function 2x. The functions
𝑥 2 + 1, 𝑥 2 − 𝜋, and 𝑥 2 + √2 are all antiderivatives of the function 2x as can be verified by
differentiation.
The process of integration reverses the process of differentiation. In differentiation, if 𝑓(𝑥) =
2𝑥 2 then 𝑓 ′ (𝑥) = 4𝑥. Thus the integral of 4x is 2𝑥 2 , i.e. integration is the process of moving
from 𝑓 ′ (𝑥) to𝑓(𝑥). By similar reasoning, the integral of 2t is𝑡 2 .
𝑑𝑦
In differentiation, the differential coefficient 𝑑𝑥 indicates that a function of x is being
differentiated with respect to x, the ‘dx’ indicating that it is ‘with respect to x’. In integration, the
variable of integration is shown by adding d (the variable) after the function to be integrated.
Thus ∫ 4𝑥 𝑑𝑥 means ‘the integral of 4x with respect to x’, and ∫ 2𝑡 𝑑𝑡 means ‘the integral of 2t
with respect to t’.
As stated above, the differential coefficient of 2𝑥 2 is 4x, hence ∫ 4𝑥𝑑𝑥 = 2𝑥 2 . However, the
differential coefficient of 2𝑥 2 + 7 is also 4x. Hence∫ 4𝑥𝑑𝑥 is also equal to 2𝑥 2 + 7. To allow for
the possible presence of a constant, whenever the process of integration is performed, a constant
‘c’ is added to the result. Thus ∫ 4𝑥 𝑑𝑥 = 2𝑥 2 + 𝑐 and ∫ 2𝑡 𝑑𝑡 = 𝑡 2 + 𝑐
‘c’ is called the arbitrary constant of integration.
3
The general solution of integrals of the form 𝒂𝒙𝒏
The general solution of integrals of the form ∫ 𝑎𝑥 𝑛 𝑑𝑥 where a and n are constants is given by:
∫ 𝒂𝒙𝒏 𝒅𝒙 = 𝒂
𝒙𝒏+𝟏
+𝒄
𝒏+𝟏
This rule is true when n is fractional, zero, or a positive or negative integer, with the exception of
𝑛 = −1.
Using this rule gives:
3𝑥 4+1
𝟑
(i)
∫ 3𝑥 4 𝑑𝑥 =
(ii)
∫ 𝑥 2 𝑑𝑥 = ∫ 2𝑥 −2 𝑑𝑥 =
(iii)
∫ √𝑥 𝑑𝑥 = ∫ 𝑥 2 𝑑𝑥 =
4+1
+ 𝑐 = 𝟓 𝒙𝟓 + 𝒄
2
1
2𝑥 −2+1
1
𝑥2
−2+1
+𝑐 =
−1
+𝑐 =
−𝟐
𝒙
+ 𝒄, and
3
+1
1
+1
2
2𝑥 −1
+𝑐 =
𝑥2
3
2
𝟐
+ 𝑐 = 𝟑 √𝒙𝟑 + 𝒄
Each of these three results may be checked by differentiation.
(a) The integral of a constant k is 𝑘𝑥 + 𝑐.
For example,
∫ 8 𝑑𝑥 = 𝟖𝒙 + 𝒄
(b) When a sum of several terms is integrated the result is the sum of the integrals of the separate
terms.
For example,
∫(3𝑥 + 2𝑥 2 − 5) 𝑑𝑥 = ∫ 3𝑥 𝑑𝑥 + ∫ 2𝑥 2 𝑑𝑥 − ∫ 5 𝑑𝑥
=
𝟑𝒙𝟐 𝟐𝒙𝟑
+
− 𝟓𝒙 + 𝒄
𝟐
𝟑
Standard integrals
Since integration is the reverse process of differentiation the standard integrals listed in the table
below may be deduced and readily checked by differentiation.
4
Table 1: Standard integrals
𝑎𝑥 𝑛+1
(ii)
∫ 𝑎𝑥 𝑛 𝑑𝑥 = 𝑛+1 + 𝑐
(𝑒𝑥𝑐𝑒𝑝𝑡 𝑤ℎ𝑒𝑛 𝑛 = −1)
1
∫ cos 𝑎𝑥 𝑑𝑥 = 𝑎 sin 𝑎𝑥 + 𝑐
(iii)
∫ sin 𝑎𝑥 𝑑𝑥 = − 𝑎 cos 𝑎𝑥 + 𝑐
(iv)
∫ 𝑠𝑒𝑐 2 𝑎𝑥 𝑑𝑥 = 𝑎 tan 𝑎𝑥 + 𝑐
(v)
∫ 𝑐𝑜𝑠𝑒𝑐 2 𝑎𝑥 𝑑𝑥 = − 𝑎 cot 𝑎𝑥 + 𝑐
(vi)
∫ sec 𝑎𝑥𝑡𝑎𝑛 𝑎𝑥 𝑑𝑥 = 𝑎 sec 𝑎𝑥 + 𝑐
(vii)
∫ 𝑐𝑜𝑠𝑒𝑐 𝑎𝑥𝑐𝑜𝑡 𝑎𝑥 𝑑𝑥 = − 𝑎 𝑐𝑜𝑠𝑒𝑐 𝑎𝑥 + 𝑐
(i)
1
1
1
1
1
1
(viii) ∫ 𝑒 𝑎𝑥 𝑑𝑥 = 𝑎 𝑒 𝑎𝑥 + 𝑐
1
(ix)
∫ 𝑥 𝑑𝑥 = ln 𝑥 + 𝑐
(x)
∫ tan 𝑎𝑥 𝑑𝑥 = 𝑎 ln|sec 𝑎𝑥| + 𝑐
(xi)
∫ cot 𝑎𝑥 𝑑𝑥 = 𝑎 ln|sin 𝑎𝑥| + 𝑐
(xii)
∫ sec 𝑎𝑥 𝑑𝑥 = 𝑎 ln|sec 𝑎𝑥 + tan 𝑎𝑥| + 𝑐
1
1
1
1
(xiii) ∫ cosec 𝑎𝑥 𝑑𝑥 = − 𝑎 ln|cosec 𝑎𝑥 + cot 𝑎𝑥| + 𝑐
1
𝑥
(xiv)
∫ √(𝑎2 −𝑥 2 ) 𝑑𝑥 = 𝑠𝑖𝑛−1 𝑎 + 𝑐
(xv)
∫ (𝒂𝟐 +𝒙𝟐 ) 𝑑𝑥 = 𝑎 𝑡𝑎𝑛−1 𝑎 + 𝑐
𝟏
1
𝑥
Example 1. Determine (a) ∫ 5𝑥 2 𝑑𝑥 (𝑏) ∫ 2𝑡 3 𝑑𝑡.
The standard integral, ∫ 𝑎𝑥 𝑛 𝑑𝑥 =
𝑎𝑥 𝑛+1
𝑛+1
+𝑐
(a) When a = 5 and n = 2 then
5𝑥 2+1
𝟓𝒙𝟑
∫ 5𝑥 𝑑𝑥 =
+𝑐 =
+𝒄
2+1
𝟑
2
(b) When a = 2 and n = 3 then
2𝑡 3+1
2𝑡 4
𝟏
∫ 2𝑡 𝑑𝑡 =
+𝑐 =
+ 𝑐 = 𝒕𝟒 + 𝒄
3+1
4
𝟐
3
Each of these results may be checked by differentiating them.
3
Example 2. Determine ∫ (4 + 7 𝑥 − 6𝑥 2 ) 𝑑𝑥.
5
3
3
∫ (4 + 7 𝑥 − 6𝑥 2 ) 𝑑𝑥 may be written as ∫ 4 𝑑𝑥 + ∫ 7 𝑥 𝑑𝑥 − ∫ 6𝑥 2 𝑑𝑥, i.e. each term is
integrated separately. (This splitting up of terms only applies, however, for addition and
subtraction.)
3 𝑥 1+1
3
Hence ∫ (4 + 7 𝑥 − 6𝑥 2 ) 𝑑𝑥 = 4𝑥 + (7)
1+1
= 𝟒𝒙 +
𝑥 2+1
3 𝑥2
− (6) 2+1 + 𝑐 = 4𝑥 + (7)
3
− (6)
𝑥3
3
+𝑐
𝟑 𝟐
𝒙 − 𝟐𝒙𝟑 + 𝒄
𝟏𝟒
Note that when an integral contains more than one term there is no need to have an arbitrary
constant for each; just a single constant at the end is sufficient.
Example3. Determine (𝑎) ∫
2𝑥 3 −3𝑥
4𝑥
𝑑𝑥 (𝑏) ∫(1 − 𝑡)2 𝑑𝑡.
(a) Rearranging into standard integral form gives:
∫
2𝑥 3 − 3𝑥
2𝑥 3 3𝑥
𝑥2 3
1 𝑥 2+1 3
𝑑𝑥 = ∫(
− ) 𝑑𝑥 = ∫( − ) 𝑑𝑥 = ( )
− 𝑥+𝑐
4𝑥
4𝑥 4𝑥
2 4
2 2+1 4
1 𝑥3 3
𝟏
𝟑
= ( ) − 𝑥 + 𝑐 = 𝒙𝟑 − 𝒙 + 𝒄
2 3 4
𝟔
𝟒
(b) Rearranging ∫(1 − 𝑡)2 𝑑𝑡 gives:
∫(1 − 2𝑡 + 𝑡 2 ) 𝑑𝑡 = 𝑡 −
2𝑡1+1 𝑡 2+1
2𝑡 2 𝑡 3
+
+𝑐 =𝑡−
+ +𝑐
1+1 2+1
2
3
𝟏
= 𝒕 − 𝒕𝟐 + 𝒕 𝟑 + 𝒄
𝟑
This problem shows that functions often have to be rearranged into the standard form of
∫ 𝑎𝑥 𝑛 𝑑𝑥 before it is possible to integrate them.
Example 4. Determine∫
3
𝑥2
𝑑𝑥.
∫
3
𝑑𝑥 = ∫ 3𝑥 −2 𝑑𝑥.
𝑥2
Using the standard integral, ∫ 𝑎𝑥 𝑛 𝑑𝑥 when a = 3 and n = – 2 gives:
∫ 3𝑥 −2 𝑑𝑥 =
3𝑥 −2+1
−𝟑
+ 𝑐 = −3𝑥 −1 + 𝑐 =
+𝒄
−2 + 1
𝒙
6
Example 5. Determine ∫ 3√𝑥 𝑑𝑥.
𝑚
𝑛
For fractional powers it is necessary to appreciate √𝑎𝑚 = 𝑎 𝑛
1
∫ 3√𝑥 𝑑𝑥 =
1
∫ 3𝑥 2 𝑑𝑥
3
3
3𝑥 2+1
3𝑥 2
=
+𝑐 =
+ 𝑐 = 2𝑥 2 + 𝑐
1
3
2+1
2
= 𝟐√𝒙𝟑 + 𝒄
Example 6. Determine ∫
−5
4
9 √𝑡 3
𝑑𝑡.
3
1
−5 −3
5 𝑡 −4+1
5 𝑡4
4
∫ 4 𝑑𝑡 = ∫ 3 𝑑𝑡 = ∫
𝑡 𝑑𝑡 = (− )
+ 𝑐 = (− ) + 𝑐
9
9 −3 + 1
9 1
9√𝑡 3
9𝑡 4
4
4
5 4 1
𝟐𝟎 𝟒
= (− ) ( ) 𝑡 4 + 𝑐 = −
√𝒕 + 𝒄
9 1
𝟗
−5
−5
Example 7. Determine ∫
∫
(1 + 𝜃)2
√𝜃
𝑑𝜃 = ∫
√𝜃
𝑑𝜃.
(1 + 2𝜃 + 𝜃 2 )
√𝜃
= ∫(𝜃
1
(1+𝜃)2
−
1
2
+
1
2𝜃 2
3
+
𝑑𝜃 = ∫(
3
𝜃 2 )𝑑𝜃
1
2𝜃
𝜃2
𝜃2
1+
1
𝜃 (−2)+1
1
+
𝜃2
1
1
1
−
1−
2−
2 + 2𝜃 2 + 𝜃 2 )𝑑𝜃
1 ) 𝑑𝜃 = ∫(𝜃
𝜃2
1
3
2𝜃 (2)+1 𝜃 (2)+1
=
+
+
+𝑐
1
1
3
−2 + 1
+
1
+
1
2
2
5
1
𝜃 2 2𝜃 2 𝜃 2
4 3 2 5
𝟒
𝟐
=
+
+
+ 𝑐 = 2𝜃 2 + 𝜃 2 + 𝜃 2 + 𝑐 = 𝟐√𝜽 + √𝜽𝟑 + √𝜽𝟓 + 𝒄
1
3
5
3
5
𝟑
𝟓
2
2
2
Example 8. Determine (𝑎) ∫ 4 cos 3𝑥 𝑑𝑥 (𝑏) ∫ 5 sin 2𝜃 𝑑𝜃.
(a) From the table of standard integrals,
1
𝟒
∫ 4 cos 4𝑥 𝑑𝑥 = (4) ( ) sin 3𝑥 + 𝑐 = 𝐬𝐢𝐧 𝟑𝒙 + 𝒄
3
𝟑
(b) From the table of standard integrals,
1
𝟓
∫ 5 sin 2𝜃 𝑑𝜃 = (5) (− ) cos 2𝜃 + 𝑐 = − 𝒄𝒐𝒔 𝟐𝜽 + 𝒄
2
𝟐
Example 9. Determine (𝑎) ∫ 7 𝑠𝑒𝑐 2 4𝑡 𝑑𝑡 (𝑏) 3 ∫ 𝑐𝑜𝑠𝑒𝑐 2 2𝜃 𝑑𝜃.
7
(a) From the table of standard integrals,
1
𝟕
∫ 7 𝑠𝑒𝑐 2 4𝑡 𝑑𝑡 = (7) ( ) tan 4𝑡 + 𝑐 = 𝐭𝐚𝐧 𝟒𝒕 + 𝒄
4
𝟒
(b) From the table of standard integrals,
1
𝟑
3 ∫ 𝑐𝑜𝑠𝑒𝑐 2 2𝜃 𝑑𝜃 = (3) (− ) cot 2𝜃 + 𝑐 = − 𝐜𝐨𝐭 𝟐𝜽 + 𝒄
2
𝟐
2
Example 10. Determine (𝑎) ∫ 5𝑒 3𝑥 𝑑𝑥 (𝑏) ∫ 3𝑒 4𝑡 𝑑𝑡.
(a) From the table of standard integrals,
1
𝟓
∫ 5𝑒 3𝑥 𝑑𝑥 = (5) ( ) 𝑒 3𝑥 + 𝑐 = 𝒆𝟑𝒙 + 𝒄
3
𝟑
2
2
2
1
1
𝟏
(b) ∫ 3𝑒 4𝑡 𝑑𝑡 = ∫ 3 𝑒 −4𝑡 𝑑𝑡 = (3) (− 4) 𝑒 −4𝑡 + 𝑐 = − 6 𝑒 −4𝑡 + 𝑐 = − 𝟔𝒆𝟒𝒕 + 𝒄
3
Example 11. Determine (𝑎) ∫ 5𝑥 𝑑𝑥 (𝑏) ∫(
3
3
1
2𝑚2 +1
) 𝑑𝑚.
𝑚
𝟑
(a) ∫ 5𝑥 𝑑𝑥 = ∫ (5) (𝑥) 𝑑𝑥 = 𝟓 𝐥𝐧 𝒙 + 𝒄 (From the table of standard integrals)
(b) ∫(
2𝑚2 +1
2𝑚2
𝑚
𝑚
) 𝑑𝑚 = ∫(
1
1
+ 𝑚) 𝑑𝑚 = ∫(2𝑚 + 𝑚) 𝑑𝑚 =
2𝑚2
2
+ ln 𝑚 + 𝑐 = 𝒎𝟐 + 𝐥𝐧 𝒎 + 𝒄
The Fundamental Theorem of Calculus
The fundamental theorem of calculus reduces the problem of integration to anti-differentiation,
i.e., finding a function 𝐹 such that𝐹 ′ = 𝑓.
Statement of the Fundamental Theorem
Theorem - Fundamental Theorem of Calculus: Suppose that the function F is differentiable
𝑏
everywhere on [a, b] and that 𝐹′ is integrable on [a, b]. Then ∫𝑎 𝐹 ′ (𝑥) 𝑑𝑥 = 𝐹(𝑏) − 𝐹(𝑎).
In other words, if 𝑓 is integrable on [a, bJ and F is an antiderivative for 𝑓, i.e., if 𝐹 ′ = 𝑓, then
𝑏
∫𝑎 𝑓(𝑥) 𝑑𝑥 = 𝐹(𝑏) − 𝐹(𝑎).
This theorem makes it easy to evaluate integrals.
Examples
𝑏
1. Using the fundamental theorem of calculus, compute∫𝑎 𝑥 2 𝑑𝑥.
8
Solution We begin by finding an antiderivative F(x) for 𝑓(𝑥) = 𝑥 2 ; from the power rule, we
1
may take 𝐹(𝑥) = 3 𝑥 3 . Now, by the fundamental theorem, we have
𝑏
𝑏
1
1
∫𝑎 𝑥 2 𝑑𝑥 = ∫𝑎 𝑓(𝑥) 𝑑𝑥 = 𝐹(𝑏) − 𝐹(𝑎) = 3 𝑏 3 − 3 𝑎3
𝑏
1
1
We conclude that ∫𝑎 𝑥 2 𝑑𝑥 = 3 𝑏 3 − 3 𝑎3 .
Expressions of the form F(b) - F(a) occur so often that it is useful to have a special notation for
𝑏
them: 𝐹(𝑥)| means F(b) - F(a). One also writes 𝐹(𝑥) = ∫ 𝑓(𝑥) 𝑑𝑥 for the antiderivative (also
𝑎
called an indefinite integral). In terms of this notation, we can write the formula of the
fundamental theorem of calculus in the form:
𝑏
𝑏
𝑏
𝑏
∫ 𝑓(𝑥) 𝑑𝑥 = 𝐹(𝑥)| 𝑜𝑟 ∫ 𝑓(𝑥) 𝑑𝑥 = (∫ 𝑓(𝑥) 𝑑𝑥)|
𝑎
𝑎
𝑎
𝑎
where F is an antiderivative of 𝑓on [a. b] .
6
2. Find ∫2 (𝑥 2 + 1)𝑑𝑥.
1
Solution By the sum and power rules for anti-derivatives, an anti-derivative for 𝑥 2 + 1 is 3 𝑥 3 +
6
1
23
6 63
𝑥. By the fundamental theorem∫2 (𝑥 2 + 1)𝑑𝑥 = ( 𝑥 3 + 𝑥)| = + 6 − ( + 2)
3
3
2 3
2
1
= 78 − 4 3 = 73 3
Exercises
1
1. Evaluate ∫0 𝑥 4 𝑑𝑥.
3
2. Find ∫0 (𝑥 2 + 3𝑥) 𝑑𝑥.
1
3. Find ∫−1(𝑡 4 + 𝑡 9 ) 𝑑𝑡.
10 𝑡 4
4. Find∫0
100
− 𝑡 2 ) 𝑑𝑡.
Definite integrals
Integrals containing an arbitrary constant c in their results are called indefinite integrals since
their precise value cannot be determined without further information. Definite integrals are
9
𝑏
those in which limits are applied. If an expression is written as [𝑋] , ‘b’ is called the upper limit
𝑎
and ‘a’ the lower limit.
𝑏
The operation of applying the limits is defined as[𝑋] = (b) − (a).
𝑎
3
The increase in the value of the integral 𝑥 2 as x increases from 1 to 3 is written as∫1 𝑥 2 𝑑𝑥.
Applying the limits gives:
3
𝑥3
33
13
1
𝟐
3
∫ 𝑥 2 𝑑𝑥 = [ + 𝑐] = ( + 𝑐) − ( + 𝑐) = (9 + 𝑐) − ( + 𝑐) = 𝟖
1
3
3
3
3
𝟑
1
Note that the ‘c’ term always cancels out when limits are applied and it need not be shown with
definite integrals.
Examples
1. Evaluate
2
3
2
3𝑥 2
(a) ∫1 3𝑥 𝑑𝑥 (𝑏) ∫−2(4 − 𝑥 2 )𝑑𝑥.
(a) ∫1 3𝑥 𝑑𝑥 = [
2
3
3
1
𝟏
2
] = {2 (2)2 } − {2 (1)2 } = 6 − 1 2 = 𝟒 𝟐
1
𝑥3
3
(b) ∫−2(4 − 𝑥 2 )𝑑𝑥 = [4𝑥 −
]
3
(−2)3
33
3
= {4(3) − 3 } − {4(−2) − 3 }
−2
= {12 − 9}— {−8 −
4 𝜃+2
2. Evaluate ∫1
4 𝜃+2
∫1 √𝜃 𝑑𝜃
=
√𝜃
4 𝜃
∫1 ( 1
𝜃2
2
[ √𝜃 3 + 4√𝜃]
3
4 𝜃
𝑑𝜃 = ∫1 (
+
2
1
𝜃2
√𝜃
+
) 𝑑𝜃 =
2
√𝜃
−8
1
𝟏
} = {3} − {−5 } = 𝟖
3
3
𝟑
), taking positive square roots only.
1
4
∫1 (𝜃 2
−
1
( +1)
𝜃 2
1
2
+ 2𝜃 ) 𝑑𝜃 = [
1
+1
2
+
−1
( )+1
2𝜃 2
−1
+1
2
4
1
2
2
= [3 √43 + 4√4] − [3 √(1)3 + 4√1]
={
16
2
1
2
𝟐
+ 8} − { + 4} = 5 + 8 − − 4 = 𝟖
3
3
3
3
𝟑
𝜋
3. Evaluate ∫02 3 sin 2𝑥 𝑑𝑥.
10
3
𝜃2
4
] =[3 +
1
2
1
2𝜃2
1
2
4
] =
1
𝜋
2
1
𝜋
𝜋
3
3
𝜋
3
∫0 3 sin 2𝑥 𝑑𝑥 = [(3) (− 2) cos 2𝑥] 2 =[− 2 cos 2𝑥] 2 ={− 2 cos 2( 2 )} − {− 2 cos 0} =
0
0
3
3
3
3
3
3
{− 2 cos 𝜋} − {− 2 cos 0} = {− 2 (−1)} − {− 2 (1)} = 2 + 2 = 𝟑
2
4. Evaluate ∫1 4 cos 3𝑡 𝑑𝑡.
2
1
4
4
4
2
2
∫ 4 cos 3𝑡 𝑑𝑡 = [4( )sin 3𝑡] = [ sin 3𝑡] = { sin 6} − { sin 3}
1
1
3
3
3
3
1
Note that limits of trigonometric functions are always expressed in radians—thus, for example,
sin 6 means the sine of 6 radians = −0.279415...
2
4
4
Hence ∫1 4 cos 3𝑡 𝑑𝑡 = {3 (−0.279415} − {3 (0.141120)} = (−0.37255) − (0.18816)
= −𝟎. 𝟓𝟔𝟎𝟕
5. Evaluate
2
4 3
(a) ∫1 4𝑒 2𝑥 𝑑𝑥 (𝑏) ∫1
4𝑢
𝑑𝑢, each correct to 4 significant figures.
2
4
2
2
(a) ∫1 4𝑒 2𝑥 𝑑𝑥 = [2 𝑒 2𝑥 ] = [2𝑒 2𝑥 ] = 2(𝑒 4 − 𝑒 2 ) = 2(54.5982 − 7.3891) = 𝟗𝟒. 𝟒𝟐
1
1
3
3
4 3
(1.3863 − 0) = 𝟏. 𝟎𝟒𝟎
𝑑𝑢
=
[
ln
𝑢]
=
(ln
4
−
ln
1)
=
4𝑢
4
4
1 4
4 3
(b) ∫1
Techniques of integration
Integration is a reverse of the process of differentiation. Given a derived function
𝑑𝑦
= 𝑓(𝑥)
𝑑𝑥
𝑑𝑦 = 𝑓(𝑥)𝑑𝑥
Integrating both sides, ∫ 𝑑𝑦 = ∫ 𝑓(𝑥) 𝑑𝑥
𝑦(𝑥) = ∫ 𝑓(𝑥) 𝑑𝑥 + 𝑐
Functions which require integrating are not always in the ‘standard form’ shown in Table 1.
However, it is often possible to change a function into a form which can be integrated by using
either:
(i) an algebraic substitution,
11
(ii) a trigonometric or hyperbolic substitution,
(iii) partial fractions,
(iv) the t = tanθ/2 substitution,
(v) integration by parts, or
(vi) reduction formulae.
Power rule
∫ 𝑎𝑥 𝑛 𝑑𝑥 =
𝑎𝑥 𝑛+1
𝑛+1
+ 𝑐, 𝑎 being a constant.
Examples
1.∫
𝑥 6 −2𝑥+4
𝑥4
2.∫(√𝑥 −
𝑑𝑥 = ∫(𝑥 2 − 2𝑥 −3 + 4𝑥 −4 ) 𝑑𝑥 =
4
3 √𝑥 5 ) 𝑑𝑥
1
2
𝑥3
3
−
1
5
4
= ∫(𝑥 − 3𝑥 ) 𝑑𝑥 =
1
𝑥 2
1
1
2
2𝑥 −2
−2
+
4𝑥 −5
−5
9
−
3𝑥 4
9
4
𝟐
𝟑
+𝑐 =
𝟒
𝒙𝟑
𝟑
𝟒
+ 𝒙−𝟐 − 𝟑 𝒙−𝟑 + 𝒄
𝟗
+ 𝑐 = 𝟑 𝒙𝟐 − 𝟗 𝒙𝟒 + 𝒄
Exercise
1.∫(𝑥 3 + 3)2 𝑑𝑥
3
2.(
√𝑥−6√𝑥 3 +1
)𝑑𝑥
𝑥3
Substitution Technique
In this method, the substitution usually made is to let u be equal to some function f(x) such that
the integrand is reduced to a standard integral.
It is found that integrals of the forms,
𝑓 ′ (𝑥)
𝑘 ∫[𝑓(𝑥)] 𝑓 (𝑥) 𝑑𝑥 𝑎𝑛𝑑 𝑘 ∫
𝑑𝑥
[𝑓(𝑥)]𝑛
𝑛 ′
(where k and n are constants) can both be integrated by substituting u for f(x).
Examples on integration using substitution
1. Determine ∫ cos(3x + 7) 𝑑𝑥.
12
∫ cos(3x + 7) 𝑑𝑥 is not a standard integral of the form shown in Table 1, thus an algebraic
substitution is made.
𝑑𝑢
Let u=3x+7 then 𝑑𝑥 = 3 and rearranging gives 𝑑𝑥 =
∫ cos(3x + 7) 𝑑𝑥 = ∫ (cos 𝑢)
𝑑𝑢
3
𝑑𝑢
3
. Hence,
1
= ∫ 3 cos 𝑢 𝑑𝑢, which is a standard integral
1
= 3 sin 𝑢 + 𝑐
Rewriting u as (3x+7) gives:
∫ cos(3x + 7) 𝑑𝑥 =
𝟏
𝐬𝐢𝐧(𝟑𝒙 + 𝟕) + 𝒄,
𝟑
which may be checked by differentiating it.
2. Determine ∫ x cos(3x 2 − 4) 𝑑𝑥.
Let 𝑢 = 3x 2 − 4 then
du
dx
= 6x and dx =
du
6x
du
1
1
𝟏
Hence ∫ x cos(3x 2 − 4) 𝑑𝑥 = ∫ 𝑥𝑐𝑜𝑠 𝑢 6x = 6 ∫ cos u du = 6 sin u + c = 𝟔 𝐬𝐢𝐧(𝟑𝐱 𝟐 − 𝟒) + 𝐜
Practice
Determine (𝑎) ∫ cos(4x + 5) 𝑑𝑥
(𝑏) ∫(𝑥 2 + 1) cos(3𝑥 3 + 9𝑥 − 5) 𝑑𝑥 (𝐻𝑖𝑛𝑡: 𝑙𝑒𝑡 𝑢 = 3𝑥 3 + 9𝑥 − 5)
(c) ∫
cos √𝑥
√𝑥
𝑑𝑥 (𝐿𝑒𝑡 𝑢 = √𝑥) (𝑑) ∫ 𝑒 −3𝑥 cos(4 + 𝑒 −3𝑥 ) 𝑑𝑥 (𝐿𝑒𝑡 𝑢 = 4 + 𝑒 −3𝑥 )
3.Determine ∫ sin 6𝑥 𝑑𝑥.
𝑑𝑢
Let u = 6x, then 𝑑𝑥 = 6 𝑎𝑛𝑑 𝑑𝑥 =
Thus ∫ sin 6𝑥 𝑑𝑥 = ∫ sin 𝑢
𝑑𝑢
6
𝑑𝑢
6
1
1
4. Determine ∫ 𝑥𝑠𝑖𝑛 (6 − 3𝑥 2 ) 𝑑𝑥.
𝑑𝑢
𝟏
= 6 ∫ sin 𝑢 𝑑𝑢 = − 6 cos 𝑢 + 𝑐 = − 𝟔 𝐜𝐨𝐬 𝟔𝒙 + 𝒄
𝑑𝑢
Let u = 6 − 3𝑥 2 , then 𝑑𝑥 = −6𝑥 𝑎𝑛𝑑 𝑑𝑥 = −6𝑥
13
𝑑𝑢
1
Hence ∫ 𝑥𝑠𝑖𝑛 (6 − 3𝑥 2 ) 𝑑𝑥 = ∫ 𝑥𝑠𝑖𝑛 𝑢 −6𝑥 = 6 ∫ −sin 𝑢 𝑑𝑢
=
1
𝟏
cos 𝑢 + 𝑐 = 𝐜𝐨𝐬(𝟔 − 𝟑𝒙𝟐 ) + 𝒄
6
𝟔
Practice
Determine (𝑎) ∫(𝜋𝑥 + 1) sin(𝜋𝑥 2 + 2𝑥 − 5) 𝑑𝑥 (𝐿𝑒𝑡 𝑢 = 𝜋𝑥 2 + 2𝑥 − 5)
(b) ∫(𝑥 2 + 1) sin(𝑥 3 + 3𝑥 − 5) 𝑑𝑥 (𝐿𝑒𝑡 𝑢 = 𝑥 3 + 3𝑥 − 5)
(c) ∫ 𝑒 2𝑥 sin(5 − 𝑒 2𝑥 ) 𝑑𝑥 (𝐿𝑒𝑡 𝑢 = 5 − 𝑒 2𝑥 )
5. Find ∫ 𝑥(1 + 3𝑥)5 𝑑𝑥.
Let 𝑢 = 1 + 3𝑥 … (𝑎)
𝑑𝑢
𝑑𝑢
= 3, ℎ𝑒𝑛𝑐𝑒
= 𝑑𝑥
𝑑𝑥
3
Substituting into the question, ∫ 𝑥(1 + 3𝑥)5 𝑑𝑥 = ∫ 𝑥(𝑢)5
𝑑𝑢
3
1
𝑑𝑢 = ∫ 3 𝑥𝑢5 𝑑𝑢
From equation (a), 𝑢 = 1 + 3𝑥;
Thus 𝑥 =
𝑢−1
3
1
1 𝑢−1
𝑎𝑛𝑑 ℎ𝑒𝑛𝑐𝑒 ∫ 3 𝑥𝑢5 𝑑𝑢 = ∫ 3 (
3
1 𝑢6
) 𝑢5 𝑑𝑢 = ∫ 3 ( 3 −
𝑢5
3
) 𝑑𝑢
1 1 𝑢7
1 𝑢6
1 1
1
= { ( ) − ( )} + 𝑐 = { (1 + 3𝑥)7 − (1 + 3𝑥)6 } + 𝑐
3 3 7
3 6
3 21
18
=
𝟏
𝟏
(𝟏 + 𝟑𝒙)𝟕 −
(𝟏 + 𝟑𝒙)𝟔 + 𝒄
𝟔𝟑
𝟓𝟒
Practice
Find ∫ 𝑥√2𝑥 + 1 𝑑𝑥
6.Find ∫(2𝑥 − 5)7 𝑑𝑥.
(2x−5) may be multiplied by itself 7 times and then each term of the result integrated. However,
this would be a lengthy process, and thus an algebraic substitution is made.
𝑑𝑢
Let u = (2x−5) then 𝑑𝑥 = 2 𝑎𝑛𝑑 𝑑𝑥 =
𝑑𝑢
2
14
Hence∫(2𝑥 − 5)7 𝑑𝑥 = ∫ 𝑢7
𝑑𝑢
2
1 𝑢8
1
1
= 2 ∫ 𝑢7 𝑑𝑢 = 2 ( 8 ) + 𝑐 = 16 𝑢8 + 𝑐
Rewriting u as (2x−5) gives:
∫(2𝑥 − 5)7 𝑑𝑥 =
𝟏
(𝟐𝒙 − 𝟓)𝟖 + 𝒄
𝟏𝟔
4
7.Find ∫ (5𝑥−3) 𝑑𝑥.
𝑑𝑢
Let u = 5x – 3, then 𝑑𝑥 = 5 𝑎𝑛𝑑 𝑑𝑥 =
4
4 𝑑𝑢
Hence ∫ (5𝑥−3) 𝑑𝑥 = ∫ 𝑢
5
4
𝑑𝑢
1
5
4
𝟒
= 5 ∫ 𝑢 𝑑𝑢 = 5 ln 𝑢 + 𝑐 = 𝟓 𝐥𝐧(𝟓𝒙 − 𝟑) + 𝒄
Practice
𝑥
(i) Determine (𝑎) ∫ 2+3𝑥 2 𝑑𝑥 (𝐿𝑒𝑡 𝑢 = 2 + 3𝑥 2 ) (𝑏) ∫
2𝑥
√(4𝑥 2 −1)
𝑑𝑥 (𝐿𝑒𝑡 𝑢 = 4𝑥 2 − 1)
𝟏
𝟏
𝐴𝑛𝑠𝑤𝑒𝑟: 𝐥𝐧(𝟐 + 𝟑𝒙𝟐 ) + 𝒄 ; √(𝟒𝒙𝟐 − 𝟏) + 𝒄
𝟔
𝟐
𝑥3
𝑥+1
(ii) Determine (𝑎) ∫ 5+𝑥 4 𝑑𝑥 (𝐿𝑒𝑡 𝑢 = 5 + 𝑥 4 ) (𝑏) ∫ 𝑥 2 +2𝑥−5 𝑑𝑥(𝐿𝑒𝑡 𝑢 = 𝑥 2 + 2𝑥 − 5)
𝑒 3𝑥
𝑥
(c) ∫ 1−3𝑥 2 𝑑𝑥 (𝑙𝑒𝑡 𝑢 = 1 − 3𝑥 2 ) (d) ∫ 2+𝑒 3𝑥 𝑑𝑥(𝐻𝑖𝑛𝑡: 𝑙𝑒𝑡 𝑢 = 2 + 𝑒 3𝑥 )
𝑠𝑒𝑐 2 2𝑥
cos 3𝑥
(e) ∫ 1+sin 3𝑥 𝑑𝑥 (𝐻𝑖𝑛𝑡: 𝑙𝑒𝑡 𝑢 = 1 + sin 3𝑥) (𝑓) ∫ 1+tan 2𝑥 𝑑𝑥 (𝐻𝑖𝑛𝑡: 𝑙𝑒𝑡 𝑢 = 1 + tan 2𝑥)
1
8.Evaluate ∫0 2𝑒 6𝑥−1 𝑑𝑥, correct to 4 significant figures.
𝑑𝑢
Let u = 6x−1 then 𝑑𝑥 = 6 𝑎𝑛𝑑 𝑑𝑥 =
Hence ∫ 2𝑒 6𝑥−1 𝑑𝑥 = ∫ 2𝑒 𝑢
𝑑𝑢
6
1
𝑑𝑢
6
1
1
= ∫ 𝑒 𝑢 𝑑𝑢 = 𝑒 𝑢 + 𝑐 = 𝑒 6𝑥−1 + 𝑐
3
3
3
Thus
1
1
1 1
∫0 2𝑒 6𝑥−1 𝑑𝑥 = 3 [𝑒 6𝑥−1 ] = 3 [𝑒 5 − 𝑒 −1 ] = 𝟒𝟗. 𝟑𝟓,
0
correct to 4 significant figures.
Practice
15
Determine (𝑎) ∫ 𝑒 3𝑥 𝑑𝑥 (𝑏) ∫ 𝑥 2 𝑒 2𝑥
3 −4
𝑑𝑥 (𝐻𝑖𝑛𝑡: 𝐿𝑒𝑡 𝑢 = 2𝑥 3 − 4)
(𝑐) ∫ cos 4𝑥𝑒 sin 4𝑥 𝑑𝑥 (𝐻𝑖𝑛𝑡: 𝐿𝑒𝑡 𝑢 = sin 4𝑥)
9. Determine ∫ 3𝑥(4𝑥 2 + 3)5 𝑑𝑥.
𝑑𝑢
Let u = (4𝑥 2 + 3) then 𝑑𝑥 = 8𝑥 𝑎𝑛𝑑 𝑑𝑥 =
𝑑𝑢
8𝑥
Hence
𝑑𝑢
3
∫ 3𝑥(4𝑥 2 + 3)5 𝑑𝑥 = ∫ 3𝑥(𝑢)5 8𝑥 = 8 ∫ 𝑢5 𝑑𝑢, by cancelling x.
The original variable ‘x’ has been completely removed and the integral is now only in terms of
u and is a standard integral.
3 𝑢6
3
1
𝟏
Hence 8 ∫ 𝑢5 𝑑𝑢 = 8 ( 6 ) + 𝑐 = 16 𝑢6 + 𝑐 = 𝟏𝟔 (𝟒𝒙𝟐 + 𝟑)𝟔 + 𝒄
Practice
𝑥+1
Determine (𝑎) ∫ 𝑥 2 (1 + 4𝑥 3 )3 𝑑𝑥 (𝑏) ∫ √2𝑥 2
+4𝑥−1
𝑑𝑥 (𝐻𝑖𝑛𝑡: 𝐿𝑒𝑡 𝑢 = 2𝑥 2 + 4𝑥 − 1)
𝜋
10. Evaluate ∫06 24 𝑠𝑖𝑛5 𝜃 cos 𝜃 𝑑𝜃.
𝑑𝑢
𝑑𝑢
Let u = sinθ then 𝑑𝜃 = cos 𝜃 𝑎𝑛𝑑 𝑑𝜃 = cos 𝜃
𝑑𝑢
Hence ∫ 24 𝑠𝑖𝑛5 𝜃 cos 𝜃 𝑑𝜃 = ∫ 24𝑢5 cos 𝜃 cos 𝜃 = ∫ 24𝑢5 𝑑𝑢 = 24
𝑢6
6
+ 𝑐 = 4𝑢6 + 𝑐
= 4𝑠𝑖𝑛6 𝜃 + 𝑐
𝜋
𝜋
𝜋 6
1 6
Thus ∫06 24 𝑠𝑖𝑛5 𝜃 cos 𝜃 𝑑𝜃 = [4𝑠𝑖𝑛6 𝜃] 6 = 4 {(sin 6 ) − (sin 0)6 } = 4 [(2) − 0]
0
=
𝟏
𝒐𝒓 𝟎. 𝟎𝟔𝟐𝟓
𝟏𝟔
11. Evaluate ∫ 𝑠𝑒𝑐 2 (7𝑥 − 1) 𝑑𝑥
𝑑𝑢
Let u = 7x – 1, then 𝑑𝑥 = 7 𝑎𝑛𝑑 𝑑𝑥 =
𝑑𝑢
7
16
∫ 𝑠𝑒𝑐 2 (7𝑥 − 1) 𝑑𝑥 = ∫ 𝑠𝑒𝑐 2 𝑢
𝑑𝑢 1
1
𝟏
= ∫ 𝑠𝑒𝑐 2 𝑢𝑑𝑢 = tan 𝑢 + 𝑐 = 𝐭𝐚𝐧(𝟕𝒙 − 𝟏) + 𝒄
7
7
7
𝟕
Practice
Determine ∫
𝑠𝑒𝑐 2 (√𝑥)
√𝑥
𝑑𝑥 (𝐿𝑒𝑡 𝑢 = √𝑥) 𝐴𝑛𝑠𝑤𝑒𝑟: 𝟐 𝐭𝐚𝐧 √𝒙 + 𝒄
12. Evaluate ∫ 𝑥 2 𝑐𝑜𝑠𝑒𝑐 2 (1 − 3𝑥 3 ) 𝑑𝑥
𝐿𝑒𝑡 𝑢 = 1 − 3𝑥 3 , 𝑡ℎ𝑒𝑛
𝑑𝑢
𝑑𝑢
= −9𝑥 2 𝑎𝑛𝑑 𝑑𝑥 =
𝑑𝑥
−9𝑥 2
𝑑𝑢
1
1
Hence ∫ 𝑥 2 𝑐𝑜𝑠𝑒𝑐 2 (1 − 3𝑥 3 ) 𝑑𝑥 = ∫ 𝑥 2 𝑐𝑜𝑠𝑒𝑐 2 𝑢( −9𝑥 2 ) = − 9 ∫ 𝑐𝑜𝑠𝑒𝑐 2 𝑢 𝑑𝑢 = 9 cot 𝑢 + 𝑐
𝟏
= 𝟗 𝐜𝐨𝐭(𝟏 − 𝟑𝒙𝟑 ) + 𝒄
Practice
Evaluate (𝑎) ∫ 𝑐𝑜𝑠𝑒𝑐 2 (4𝑥) 𝑑𝑥 (𝑏) ∫
𝑐𝑜𝑠𝑒𝑐 2 (√𝑥)
√𝑥
𝑑𝑥 (𝐿𝑒𝑡 𝑢 = √𝑥)
(𝑐) ∫(𝑥 + 1)𝑐𝑜𝑠𝑒𝑐 2 (2𝑥 2 + 4𝑥 − 5) 𝑑𝑥(𝐿𝑒𝑡 𝑢 = 2𝑥 2 + 4𝑥 − 5)
13(a).Show that ∫ tan 𝜃 𝑑𝜃 = ln(sec 𝜃) + 𝑐
∫ tan 𝜃 𝑑𝜃 = ∫
𝐿𝑒𝑡 𝑢 = cos 𝜃 , 𝑡ℎ𝑒𝑛
sin 𝜃
Hence ∫ cos 𝜃 𝑑𝜃 = ∫
sin 𝜃
𝑢
𝑑𝑢
(− sin 𝜃) = − ∫
1
𝑑𝑢
𝑑𝑢
= − sin 𝜃 𝑎𝑛𝑑 𝑑𝜃 =
𝑑𝜃
− sin 𝜃
𝑑𝑢
𝑢
sin 𝜃
𝑑𝜃
cos 𝜃
= − ln 𝑢 + 𝑐 = − ln(cos 𝜃) + 𝑐 = 0 − ln(cos 𝜃) +
𝑐 = ln 1 − ln(cos 𝜃) + 𝑐 = ln(cos 𝜃) + 𝑐
Hence ∫ 𝐭𝐚𝐧 𝜽 𝒅𝜽 = 𝐥𝐧|𝒔𝒆𝒄 𝜽| + 𝒄
Practice
Evaluate the indefinite integrals:
i.∫ tan(5𝑥 − 2) 𝑑𝑥,
17
ii.∫ 𝑥 2 tan(5 − 𝑥 3 ) 𝑑𝑥,
iii,∫(𝑥 2 − 1) tan(3𝑥 3 − 9𝑥 − 7) 𝑑𝑥,
iv.∫
tan √𝑥
√𝑥
𝑑𝑥.
13(b) Show that ∫ cot 𝜃 𝑑𝜃 = ln|sin 𝜃| + 𝑐
Practice
Evaluate the indefinite integrals:
i.∫ cot(5𝑥) 𝑑𝑥,
ii.∫ 𝑥 2 tan(2 − 𝑥) 𝑑𝑥,
iii,∫(𝑥 − 1) tan(𝑥 2 − 2𝑥 − 1) 𝑑𝑥,
iv.∫
cot √𝑥
√𝑥
𝑑𝑥.
14(a) Show that ∫ sec 𝑥 𝑑𝑥 = ln|sec 𝑥 + tan 𝑥| + 𝑘.
Multiply through the numerator and denominator by (sec 𝑥 + tan 𝑥). Thus
∫ sec 𝑥 𝑑𝑥 = ∫
sec 𝑥.(sec 𝑥+tan 𝑥)
sec 𝑥+tan 𝑥
𝑑𝑥….(i)
Let u = sec 𝑥 + tan 𝑥 ….(ii)
𝑑𝑢
= sec 𝑥 tan 𝑥 + 𝑠𝑒𝑐 2 𝑥
𝑑𝑥
𝑑𝑢
sec 𝑥 tan 𝑥+𝑠𝑒𝑐 2 𝑥
= 𝑑𝑥 …..(iii)
Substituting (ii) and (iii) in (i),
∫ sec 𝑥 𝑑𝑥 = ∫
sec 𝑥. (sec 𝑥 + tan 𝑥)
𝑑𝑢
1
= ∫ 𝑑𝑢 = ln|𝑢| + 𝑘
2
𝑢
sec 𝑥 tan 𝑥 + 𝑠𝑒𝑐 𝑥
𝑢
Therefore, ∫ sec 𝑥 𝑑𝑥 = ln|sec 𝑥 + tan 𝑥| + 𝑘.
Practice
Evaluate the indefinite integrals:
18
1
i.∫ 2𝑥 sec(4𝑥 2 + 5) 𝑑𝑥,
𝐴𝑛𝑠 4 ln|sec(4𝑥 2 + 5) + tan(4𝑥 2 + 5)| + 𝑘
ii.∫ 2𝑥 2 sec(4𝜋𝑥 3 + 6) 𝑑𝑥.
𝐴𝑛𝑠
1
6π
ln|sec(4𝜋𝑥 3 + 6) + tan(4𝜋𝑥 3 + 6)| + 𝑘
14(b) Show that ∫ cosec 𝑥 𝑑𝑥 = − ln|cosec 𝑥 + cot 𝑥| + 𝑘.
Multiply through the numerator and denominator by (cosec 𝑥 + cot 𝑥). Thus
∫ cosec 𝑥 𝑑𝑥 = ∫
cosec 𝑥 (cosec 𝑥+cot 𝑥.)
cosec 𝑥+cot 𝑥
𝑑𝑥….(i)
Let u = (cosec 𝑥 + cot 𝑥) ….(ii)
𝑑𝑢
= − cosec 𝑥 cot 𝑥 − 𝑐𝑜𝑠𝑒𝑐 2 𝑥
𝑑𝑥
𝑑𝑢
−cosec 𝑥 cot 𝑥−𝑐𝑜𝑠𝑒𝑐 2 𝑥
= 𝑑𝑥 …..(iii)
Substituting (ii) and (iii) in (i),
∫ cosec 𝑥 𝑑𝑥 = ∫
cosec 𝑥. (cosec 𝑥 + cot 𝑥)
𝑑𝑢
1
= ∫ − 𝑑𝑢
2
𝑢
−cosec 𝑥 cot 𝑥 − 𝑐𝑜𝑠𝑒𝑐 𝑥
𝑢
1
= − ∫ 𝑑𝑢 = −ln|𝑢| + 𝑘
𝑢
Therefore, ∫ cosec 𝑥 𝑑𝑥 = − ln|cosec 𝑥 + cot 𝑥| + 𝑘
Practice
Evaluate the indefinite integrals:
i.∫ cosec(2𝑥 + 3) 𝑑𝑥,
ii.∫(𝑥 + 2) cosec(𝑥 2 + 4𝑥 − 3) 𝑑𝑥.
Change of limits
When evaluating definite integrals involving substitutions, it is sometimes more convenient to
change the limits of the integral as shown in Problems 14 and 15 below.
3
15. Evaluate ∫1 5𝑥√(2𝑥 2 + 7)𝑑𝑥, taking positive values of square roots only.
19
𝐿𝑒𝑡 𝑢 = 2𝑥 2 + 7, 𝑡ℎ𝑒𝑛
𝑑𝑢
𝑑𝑢
= 4𝑥 𝑎𝑛𝑑 𝑑𝑥 =
𝑑𝑥
4𝑥
It is possible in this case to change the limits of integration. Thus when x = 3, u = 2(3)2 + 7 =
25 and when x = 1,u = 2(1)2 + 7 = 9.
3
25
𝑑𝑢
5
25
5
25
1
Hence ∫1 5𝑥√(2𝑥 2 + 7)𝑑𝑥 = ∫9 5𝑥 √𝑢 4𝑥 = 4 ∫9 √𝑢 𝑑𝑢 = 4 ∫9 𝑢2 𝑑𝑢
Thus the limits have been changed, and it is unnecessary to change the integral back in terms of
x.
3
Thus
3
∫1 5𝑥√(2𝑥 2
2
16. Evaluate ∫0
5 𝑢2
+ 7)𝑑𝑥 = 4 [
3𝑥
√(2𝑥 2 +1)
Let 𝑢 = 2𝑥 2 + 1, 𝑡ℎ𝑒𝑛
3
2
]
5
𝟐
25 5
25 5
= 6 [√𝑢3 ] = 6 [√253 − √93 ] = 6 (125 − 27) = 𝟖𝟏 𝟑
9
9
𝑑𝑥, taking positive values of square roots only.
𝑑𝑢
𝑑𝑥
= 4𝑥 𝑎𝑛𝑑 𝑑𝑥 =
𝑑𝑢
4𝑥
When x = 2, u = 2(2)2 + 1 = 9 and when x = 0, u = 1.
Thus
2
3𝑥
∫0 √(2𝑥 2 +1) 𝑑𝑥
=
9 3𝑥 𝑑𝑢
∫1 √𝑢 4𝑥
=
9 1
𝑑𝑢
∫
1
4
√𝑢
3
=
9 −1
𝑢 2 𝑑𝑢
∫
1
4
3
1
3 𝑢2
= 4[
1
2
9 3
] = 2 [√9 − √1] = 𝟑,
1
taking positive values of square roots only.
Powers of trigonometric functions
Integrals Involving 𝒔𝒊𝒏𝟐 𝜽, 𝒄𝒐𝒔𝟐 𝜽. 𝒕𝒂𝒏𝟐 𝜽 𝒂𝒏𝒅𝒄𝒐𝒕𝟐 𝜽
In such cases we use the identities:
(𝑎)𝑐𝑜𝑠 2 𝜃 =
1
(1 + cos 2𝜃)
2
(𝑏)𝑠𝑖𝑛2 𝜃 =
1
(1 − cos 2𝜃)
2
Notice that both are obtained by elimination from the trigonometric identities:
(𝑖)𝑐𝑜𝑠 2 𝜃 + 𝑠𝑖𝑛2 𝜃 = 1
(𝑖𝑖)𝑐𝑜𝑠 2 𝜃 − 𝑠𝑖𝑛2 𝜃 = cos 2𝜃
20
Since 1 + 𝑡𝑎𝑛2 𝜃 = 𝑠𝑒𝑐 2 𝜃 𝑎𝑛𝑑 𝑐𝑜𝑡 2 𝜃 + 1 = 𝑐𝑜𝑠𝑒𝑐 2 𝜃, then:
(𝑐)𝑡𝑎𝑛2 𝜃 = 𝑠𝑒𝑐 2 𝜃 − 1
(𝑑)𝑐𝑜𝑡 2 𝜃 = 𝑐𝑜𝑠𝑒𝑐 2 𝜃 − 1
Examples
𝜋
1. Evaluate ∫04 2𝑐𝑜𝑠 2 4𝑡 𝑑𝑡.
1
1
𝑐𝑜𝑠 2 𝜃 = (1 + cos 2𝜃) 𝑎𝑛𝑑 𝑐𝑜𝑠 2 4𝑡 = (1 + cos 8𝑡)
2
2
𝜋
4
𝜋
1
4
2
Hence, ∫0 2𝑐𝑜𝑠 4𝑡 𝑑𝑡 = 2 ∫0 2 (1 + cos 8𝑡)𝑑𝑡 = [𝑡 +
sin 8𝑡
8
𝝅
𝜋
𝜋
] 4 𝑑𝑡 = [ 4 +
0
𝜇
4
sin 8( )
sin 0
8
8
] − [0 +
𝟒
Or 0.7854
2.Determine ∫ 𝑐𝑜𝑠 4 2𝑥 𝑑𝑥.
𝑚4 = (𝑚2 )2
1
Therefore 𝑐𝑜𝑠 4 2𝑥 = (𝑐𝑜𝑠 2 2𝑥)2 = [2 (1 + 𝑐𝑜𝑠4𝑥)]
2
Now, (𝑎 + 𝑏)2 = 𝑎2 + 2𝑎𝑏 + 𝑏 2 .
2
1
1
1
1
Thus, [2 (1 + 𝑐𝑜𝑠4𝑥)] = 4 (1 + 2𝑐𝑜𝑠4𝑥 + 𝑐𝑜𝑠 2 4𝑥) = 4 {1 + 2𝑐𝑜𝑠4𝑥 + 2 (1 + cos 8𝑥)}
1
2
1
1
Therefore, ∫ 𝑐𝑜𝑠 4 2𝑥 𝑑𝑥 = 4 ∫ 𝑑𝑥 + 4 ∫ cos 4𝑥 𝑑𝑥 + 8 ∫ 𝑑𝑥 + 8 ∫ cos 8𝑥𝑑𝑥
=
𝑥 1 1
1
1 1
𝟑
𝟏
𝟏
+ ( sin 4𝑥) + 𝑥 + ( sin 8𝑥) + 𝑐 = 𝒙 + 𝐬𝐢𝐧 𝟖𝒙 +
𝐬𝐢𝐧 𝟖𝒙 + 𝒄
4 2 4
8
8 8
𝟖
𝟖
𝟔𝟒
Practice
𝜋
Evaluate ∫04 4𝑐𝑜𝑠 4 𝜃𝑑𝜃 , correct to 4 significant figures.
Answer 2.178
3. Determine ∫ 𝑠𝑖𝑛2 3𝑥 𝑑𝑥.
𝑠𝑖𝑛2 𝜃 =
1
1
(1 − cos 2𝜃) 𝑎𝑛𝑑 𝑠𝑖𝑛2 3𝑥 = (1 − cos 6𝑥)
2
2
21
]=
1
𝟏
Hence, ∫ 𝑠𝑖𝑛2 3𝑥 𝑑𝑥 = ∫ 2 (1 − cos 6𝑥) 𝑑𝑥 = 𝟐 (𝒙 −
𝐬𝐢𝐧 𝟔𝒙
𝟔
)+𝒄
4. Find ∫ 3𝑡𝑎𝑛2 4𝑥 𝑑𝑥.
𝑡𝑎𝑛2 𝜃 = 𝑠𝑒𝑐 2 𝜃 − 1
tan 4𝑥
Thus, ∫ 3𝑡𝑎𝑛2 4𝑥 𝑑𝑥 = 3 ∫(𝑠𝑒𝑐 2 4𝑥 − 1) 𝑑𝑥 = 3 (
𝜋
4
− 𝑥) + 𝑐 = 𝟑
𝐭𝐚𝐧 𝟒𝒙
𝟒
− 𝟑𝒙 + 𝒄
1
5.Evaluate ∫𝜋3 2 𝑐𝑜𝑡 2 2𝜃 𝑑𝜃.
6
𝑐𝑜𝑡 2 𝜃 = 𝑐𝑜𝑠𝑒𝑐 2 𝜃 − 1
𝜋
31
𝜋
2
6
Hence, ∫
1
𝜋
3
2
𝜋
𝜋
3
𝜋
6
1 − cot 2𝜃
2
𝑐𝑜𝑡 2𝜃 𝑑𝜃 = 2 ∫ (𝑐𝑜𝑠𝑒𝑐 2𝜃 − 1) 𝑑𝜃 = 2 [
− cot 2( )
= 2 {(
1
2
𝜋
− 3) − (
𝜋
6
− cot 2( )
2
2
− 𝜃] 𝜋3
6
𝜋
− 6 )} =[(0.2887−1.0472) − (−0.2887−0.5236)]
= 0.0269
Exercise
Evaluate the following integrals:
1.∫ 𝑠𝑖𝑛2 2𝑥 𝑑𝑥,
2.∫ 3𝑐𝑜𝑠 2 3𝑥 𝑑,
3.∫ 5𝑡𝑎𝑛2 2𝑡 𝑑𝑡,
𝜋
4.∫04 𝑐𝑜𝑠 2 4𝑥 𝑑𝑥,
𝜋
3
𝜋
6
5.∫ 𝑐𝑜𝑡 2 𝜃 𝑑𝜃.
More examples on powers of sines and cosines
6. Determine ∫ 𝑠𝑖𝑛5 𝜃 𝑑𝜃.
Since 𝑐𝑜𝑠 2 𝜃 + 𝑠𝑖𝑛2 𝜃 = 1 then 𝑠𝑖𝑛2 𝜃 = 1 − 𝑐𝑜𝑠 2 𝜃.
Hence∫ 𝑠𝑖𝑛5 𝜃 𝑑𝜃 = ∫ sin 𝜃(𝑠𝑖𝑛2 𝜃)2 𝑑𝜃 = ∫ sin 𝜃(1 − 𝑐𝑜𝑠 2 𝜃)2 𝑑𝜃
22
= ∫ sin 𝜃(1 − 2𝑐𝑜𝑠 2 𝜃 + 𝑐𝑜𝑠 4 𝜃)𝑑𝜃 = ∫(sin 𝜃 − 2 sin 𝜃𝑐𝑜𝑠 2 𝜃 + sin 𝜃𝑐𝑜𝑠 4 𝜃)𝑑𝜃
= − 𝐜𝐨𝐬 𝜽 +
𝟐𝒄𝒐𝒔𝟑 𝜽 𝒄𝒐𝒔𝟓 𝜽
−
+𝒄
𝟑
𝟓
Whenever a power of a cosine is multiplied by a sine of power 1, or vice-versa, the integral may
be determined by inspection as shown above.
𝒄𝒐𝒔𝒏+𝟏 𝜽
In general, ∫ 𝑐𝑜𝑠 𝑛 𝜃 sin 𝜃 𝑑𝜃 = −
𝒏+𝟏
+ 𝒄 𝑎𝑛𝑑 ∫ 𝑠𝑖𝑛𝑛 𝜃 cos 𝜃 𝑑𝜃 =
𝒔𝒊𝒏𝒏+𝟏 𝜽
𝒏+𝟏
+ 𝒄.
𝜋
7. Evaluate ∫02 𝑠𝑖𝑛2 𝑥𝑐𝑜𝑠 3 𝑥 𝑑𝑥.
𝜋
2
2
𝜋
2
3
𝜋
2
∫ 𝑠𝑖𝑛 𝑥𝑐𝑜𝑠 𝑥 𝑑𝑥 = ∫ 𝑠𝑖𝑛 𝑥𝑐𝑜𝑠 𝑥𝑐𝑜𝑠 𝑥 𝑑𝑥 ∫ 𝑠𝑖𝑛2 𝑥(1 − 𝑠𝑖𝑛2 𝑥) cos 𝑥 𝑑𝑥
0
2
2
0
=∫
𝜋
2
(𝑠𝑖𝑛2
0
=[
0
𝜋
𝑠𝑖𝑛3 𝑥 𝑠𝑖𝑛5 𝑥
𝑥𝑐𝑜𝑠𝑥 − 𝑠𝑖𝑛 𝑥𝑐𝑜𝑠 𝑥) 𝑑𝑥 = [
−
]2
3
5 0
4
𝜋 3
(sin 2)
3
−
𝜋 5
(sin 2)
5
] − [0 − 0] =
1 1
𝟐
− =
3 5 𝟏𝟓
Or 0.1333
8. Find ∫ 𝑠𝑖𝑛2 𝑡𝑐𝑜𝑠 4 𝑡 𝑑𝑡.
∫ 𝑠𝑖𝑛2 𝑡𝑐𝑜𝑠 4 𝑡 𝑑𝑡 = ∫ 𝑠𝑖𝑛2 𝑡(𝑐𝑜𝑠 2 𝑡)2 𝑑𝑡 = ∫(
=
=
1 − cos 2𝑡 1 + cos 2𝑡 2
)(
) 𝑑𝑡
2
2
1
∫(1 − cos 2𝑡)( 1 + 2 cos 2𝑡 + 𝑐𝑜𝑠 2 2𝑡) 𝑑𝑡
8
1
∫(1 + 2 cos 2𝑡 + 𝑐𝑜𝑠 2 2𝑡 − cos 2𝑡 − 2𝑐𝑜𝑠 2 2𝑡 − 𝑐𝑜𝑠 3 2𝑡) 𝑑𝑡
8
1
= ∫(1 + cos 2𝑡 − 𝑐𝑜𝑠 2 2𝑡 − 𝑐𝑜𝑠 3 2𝑡) 𝑑𝑡
8
1
1 + cos 4𝑡
= ∫(1 + cos 2𝑡 − (
) − cos 2𝑡(1 − 𝑠𝑖𝑛2 2𝑡) 𝑑𝑡
8
2
23
1 1 cos 4𝑡
𝟏 𝒕 𝐬𝐢𝐧 𝟒𝒕 𝒔𝒊𝒏𝟑 𝟐𝒕
2
= ∫( −
+ cos 2𝑡𝑠𝑖𝑛 2𝑡) 𝑑𝑡 = ( −
+
)+𝒄
8 2
2
𝟖 𝟐
𝟖
𝟔
Exercise
Integrate each of the following functions with respect to the variable
1.𝑠𝑖𝑛3 𝜃
2.2𝑐𝑜𝑠 3 2𝑥
3.2𝑠𝑖𝑛3 𝑡𝑐𝑜𝑠 3 𝑡
4. 𝑠𝑖𝑛3 𝑥𝑐𝑜𝑠 2 𝑥
Products of sines and cosines
sin(𝐴 + 𝐵) = sin 𝐴 cos 𝐵 + cos 𝐴 sin 𝐵 … . (1)
sin(𝐴 − 𝐵) = sin 𝐴 cos 𝐵 − cos 𝐴 sin 𝐵 … . (2)
𝟏
Adding (1) and (2) gives 𝐬𝐢𝐧 𝑨 𝐜𝐨𝐬 𝑩 = 𝟐 [𝐬𝐢𝐧(𝑨 + 𝑩) + 𝐬𝐢𝐧(𝑨 − 𝑩)].
𝟏
Subtracting (2) from (1) gives𝐜𝐨𝐬 𝑨 𝐬𝐢𝐧 𝑩 = 𝟐 [𝐬𝐢𝐧(𝑨 + 𝑩) − 𝐬𝐢𝐧(𝑨 − 𝑩)].
Thus a product of sine and cosine can be replaced by the respective right hand side.
Similarly,
cos(𝐴 + 𝐵) = cos 𝐴 cos 𝐵 − sin 𝐴 sin 𝐵 … . (3)
cos(𝐴 − 𝐵) = cos 𝐴 cos 𝐵 + sin 𝐴 sin 𝐵 … . (4)
𝟏
Adding (3) and (4) gives 𝐜𝐨𝐬 𝑨 𝐜𝐨𝐬 𝑩 = 𝟐 [𝐜𝐨𝐬(𝑨 + 𝑩) + 𝐜𝐨𝐬(𝑨 − 𝑩)], thus a product of
cosines can be replaced by the right hand side.
𝟏
Subtracting (4) from (3) gives 𝐬𝐢𝐧 𝑨 𝐬𝐢𝐧 𝑩 = − 𝟐 [𝐜𝐨𝐬(𝑨 + 𝑩) − 𝐜𝐨𝐬(𝑨 − 𝑩)], thus a product of
sines can be replaced by the right hand side.
Examples
1.Determine ∫ sin 3𝑡 cos 2𝑡 𝑑𝑡.
24
1
1
∫ sin 3𝑡 cos 2𝑡 𝑑𝑡 = ∫ [sin(3𝑡 + 2𝑡) + sin(3𝑡 − 2𝑡)] 𝑑𝑡 = ∫ (sin 5𝑡 + sin 𝑡) 𝑑𝑡
2
2
=
𝟏 − 𝐜𝐨𝐬 𝟓𝒕
(
− 𝐜𝐨𝐬 𝒕) + 𝒄
𝟐
𝟓
1
2.Find∫ 3 (cos 5𝑥 sin 2𝑥) 𝑑𝑥.
1
1 1
1
∫ (cos 5𝑥 sin 2𝑥) 𝑑𝑥 = ∫ [sin(5𝑥 + 2𝑥) − sin(5𝑥 − 2𝑥)] 𝑑𝑥 = ∫ (sin 7𝑥 − sin 3𝑥) 𝑑𝑥
3
3 2
6
=
𝟏 − 𝐜𝐨𝐬 𝟕𝒙 𝐜𝐨𝐬 𝟑𝒙
(
+
)+𝒄
𝟔
𝟕
𝟑
1
3.Evaluate∫0 2 cos 6𝜃 cos 𝜃 𝑑𝜃, correct to 4 decimal places.
1
1
∫ 2 cos 6𝜃 cos 𝜃 𝑑𝜃 = 2 ∫
0
= [(
0
sin 7𝜃
7
+
sin 5𝜃
5
1
1
[cos(6𝜃 + 𝜃) + cos(6𝜃 − 𝜃)] 𝑑𝜃 = ∫ (cos 7𝜃 + cos 5𝜃) 𝑑𝜃
2
0
sin 7
sin 5
sin 0
sin 0
sin 7
sin 5
1
)] = [ 7 + 5 ] − [ 7 + 5 ] = [ 7 + 5 ].
0
‘sin 7’ means ‘the sine of 7 radians’ and sin 5 ‘the sine of 5 radians’.
1
Hence ∫0 2 cos 6𝜃 cos 𝜃 𝑑𝜃 = (0.09386 + (−0.19178)) − (0)
= −0.0979, correct to 4 decimal places.
4.Find 3 ∫ sin 5𝑥 sin 3𝑥 𝑑𝑥.
1
3 ∫ sin 5𝑥 sin 3𝑥 𝑑𝑥 = 3 ∫ − [cos(5𝑥 + 3𝑥) − cos(5𝑥 − 3𝑥)] 𝑑𝑥
2
3
3 sin 8𝑥 sin 2𝑥
= − ∫(cos 8𝑥 − cos 2𝑥) 𝑑𝑥 = − (
−
)+𝑐
2
2
8
2
𝟑
=
(𝟒 𝐬𝐢𝐧 𝟐𝒙 − 𝐬𝐢𝐧 𝟖𝒙) + 𝒄
𝟏𝟔
Exercise
Integrate each of the following functions with respect to the variable
1.sin 5t cos 2t.
2.2 sin 3x sin x.
25
3.3 cos 6x cos x.
1
4. 2 cos 4𝜃 sin 2𝜃.
Integration using 𝐬𝐢𝐧 𝜽 substitution
Examples
1.Determine∫
1
√(𝑎2 −𝑥 2 )
Let 𝑥 = asin 𝜃, 𝑡ℎ𝑒𝑛
Hence ∫
1
√(𝑎2 −𝑥 2 )
𝑑𝑥
𝑑𝑥
𝑑𝜃
= acos 𝜃 𝑎𝑛𝑑 𝑑𝑥 = acos 𝜃𝑑𝜃.
𝑑𝑥 = ∫
1
√(𝑎2 −𝑎2 𝑠𝑖𝑛2 𝜃)
acos 𝜃𝑑𝜃 = ∫
acos 𝜃𝑑𝜃
[√𝑎2 (1−𝑠𝑖𝑛2 𝜃)]
=∫
acos 𝜃𝑑𝜃
,
√(𝑎2 𝑐𝑜𝑠2 𝜃)
𝑠𝑖𝑛𝑐𝑒 𝑐𝑜𝑠 2 𝜃 + 𝑠𝑖𝑛2 𝜃 = 1.
=∫
acos 𝜃𝑑𝜃
= ∫ 𝑑𝜃 = 𝜃 + 𝑐
𝑎𝑐𝑜𝑠𝜃
𝑥
𝑥
Since 𝑥 = asin 𝜃, 𝑡ℎ𝑒𝑛 sin 𝜃 = 𝑎 𝑎𝑛𝑑 𝜃 = 𝑠𝑖𝑛−1 𝑎.
Hence ∫
1
𝒙
√(𝑎2 −𝑥 2 )
3
2. Evaluate ∫0
𝑑𝑥 = 𝒔𝒊𝒏−𝟏 𝒂 + 𝒄.
1
√(9−𝑥 2 )
𝑑𝑥.
3
From the above example, ∫0
1
√(9−𝑥 2 )
𝒙 3
𝑑𝑥 = [𝒔𝒊𝒏−𝟏 𝟑] , 𝑠𝑖𝑛𝑐𝑒 𝑎 = 3.
0
= (𝑠𝑖𝑛−1 1 − 𝑠𝑖𝑛−1 0) =
𝜋
𝝅
− 0 = 𝒐𝒓 𝟏. 𝟓𝟕𝟎𝟖.
2
𝟐
3.Find ∫ √(𝑎2 − 𝑥 2 ) 𝑑𝑥.
Let 𝑥 = asin 𝜃, 𝑡ℎ𝑒𝑛
𝑑𝑥
𝑑𝜃
= acos 𝜃 𝑎𝑛𝑑 𝑑𝑥 = acos 𝜃𝑑𝜃.
Hence ∫ √(𝑎2 − 𝑥 2 ) 𝑑𝑥 = ∫ √(𝑎2 − 𝑎2 𝑠𝑖𝑛2 𝜃) acos 𝜃𝑑𝜃 = ∫ √(𝑎2 (1 − 𝑠𝑖𝑛2 𝜃) acos 𝜃𝑑𝜃 =
26
= ∫ √(𝑎2 (𝑐𝑜𝑠 2 𝜃) acos 𝜃𝑑𝜃
= ∫(acos 𝜃) (acos 𝜃𝑑𝜃) = 𝑎2 ∫ 𝑐𝑜𝑠 2 𝜃 𝑑𝜃
= 𝑎2 ∫(
=
1 + cos 2𝜃
)𝑑𝜃 , 𝑠𝑖𝑛𝑐𝑒 (cos 2𝜃 = 2𝑐𝑜𝑠 2 𝜃 − 1)
2
𝑎2
sin 2𝜃
𝑎2
2 sin 𝜃 cos 𝜃
𝑎2
(𝜃 +
) + 𝑐 = (𝜃 +
) + 𝑐 = (𝜃 + sin 𝜃 cos 𝜃) + 𝑐
2
2
2
2
2
𝑥
𝑥
Since 𝑥 = asin 𝜃, 𝑡ℎ𝑒𝑛 sin 𝜃 = 𝑎 𝑎𝑛𝑑 𝜃 = 𝑠𝑖𝑛−1 𝑎.
𝑥
Also, 𝑐𝑜𝑠 2 𝜃 + 𝑠𝑖𝑛2 𝜃 = 1, 𝑓𝑟𝑜𝑚 𝑤ℎ𝑖𝑐ℎ cos 𝜃 = √1 − 𝑠𝑖𝑛2 𝜃 = √[1 − (𝑎)2 ] = √(
𝑎2 −𝑥2
√(𝑎2 − 𝑥 2 )
=
𝑎
Thus ∫ √(𝑎2 − 𝑥 2 ) 𝑑𝑥 =
𝑎2
2
(𝜃 + sin 𝜃 cos 𝜃) + 𝑐 =
=
𝑎2
2
𝑥
𝑥 √(𝑎2 −𝑥 2 )
[𝑠𝑖𝑛−1 𝑎 + (𝑎)
𝑎
𝒂𝟐
𝒙 𝒙
𝒔𝒊𝒏−𝟏 + √(𝒂𝟐 − 𝒙𝟐 ) + 𝒄
𝟐
𝒂 𝟐
4
4.Evaluate ∫0 √(16 − 𝑥 2 ) 𝑑𝑥
4
𝟏𝟔
𝒙
𝒙
From above example, ∫0 √(16 − 𝑥 2 ) 𝑑𝑥 = [ 𝟐 𝒔𝒊𝒏−𝟏 𝟒 + 𝟐 √(𝟏𝟔 − 𝒙𝟐 )]
𝜋
4
0
= [8𝑠𝑖𝑛−1 1 + 2√(0)] − [8𝑠𝑖𝑛−1 0 + 0] = 8𝑠𝑖𝑛−1 1 = 8 ( 2 ) = 𝟒𝝅 or 12.57
Exercise
5
1.Determine ∫ √4−𝑡 2 𝑑𝑡.
3
2.Determine ∫ √9−𝑥 2 𝑑𝑥.
3.Determine ∫ √4 − 𝑥 2 𝑑𝑥.
4. Determine ∫ √16 − 9𝑡 2 𝑑𝑥.
4
5.Evaluate ∫0 √16 − 𝑥 2 𝑑𝑥.
Integration using 𝒕𝒂𝒏𝜽 substitution
27
]+𝑐
𝑎2
)
Examples
1
1.Determine∫ (𝑎2 +𝑥 2 ) 𝑑𝑥.
𝑑𝑥
Let 𝑥 = atan 𝜃 𝑡ℎ𝑒𝑛
𝑑𝜃
= 𝑎𝑠𝑒𝑐 2 𝜃 𝑎𝑛𝑑 𝑑𝑥 = 𝑎𝑠𝑒𝑐 2 𝜃𝑑𝜃.
1
𝑎𝑠𝑒𝑐 2 𝜃𝑑𝜃
1
Hence ∫ (𝑎2 +𝑥 2 ) 𝑑𝑥 = ∫ (𝑎2 +𝑎2 𝑡𝑎𝑛2 𝜃) (𝑎𝑠𝑒𝑐 2 𝜃𝑑𝜃) = ∫ 𝑎2 (1+𝑡𝑎𝑛2 𝜃) = ∫
𝑎𝑠𝑒𝑐 2 𝜃𝑑𝜃
𝑎2 𝑠𝑒𝑐 2 𝜃
,
𝑠𝑖𝑛𝑐𝑒 1 + 𝑡𝑎𝑛2 𝜃 = 𝑠𝑒𝑐 2 𝜃.
1
1
= ∫ 𝑑𝜃 = 𝜃 + 𝑐.
𝑎
𝑎
𝑥
Since 𝑥 = atan 𝜃, 𝜃 = 𝑡𝑎𝑛−1 𝑎.
𝟏
𝟏
𝒙
Hence ∫ (𝒂𝟐 +𝒙𝟐 ) 𝒅𝒙 = 𝒂 𝒕𝒂𝒏−𝟏 𝒂 + 𝒄.
2
2. Evaluate ∫0
1
𝑑𝑥
(4+𝑥 2 )
2
From the above example, ∫0
1
1
(4+𝑥 2 )
1
𝑥 2
𝑑𝑥 = 2 [𝑡𝑎𝑛−1 2] since a = 2.
0
1 𝜋
= 2(𝑡𝑎𝑛−1 1 − 𝑡𝑎𝑛−1 0) = 2 (4 − 0) =
1
3.Evaluate ∫0
1
∫0
5
(3+2𝑥 2 )
𝑑𝑥 = ∫0
5
5
3
2
2[ +𝑥 2 ]
=
5
2
𝟖
𝒐𝒓 𝟎. 𝟑𝟗𝟐𝟕
𝑑𝑥, correct to 4 decimal places.
1
5
(3+2𝑥 2 )
𝝅
1
𝑑𝑥 = 2 ∫0
5
2
1
5
3
[(√ )2 +𝑥 2 ]
2
1
√3
( 2) [
𝑡𝑎𝑛−1
𝑑𝑥 = 2 [
1
3
√
2
𝑡𝑎𝑛−1
𝑥
3
√
2
3
1
] , since 𝑎 = √2
0
𝑥 1 5 2
2 1
= √ [𝑡𝑎𝑛−1 √ 𝑥]
0 2 3
3 0
√3
2]
2
= 2 √(3) [𝑡𝑎𝑛−1 √(3) − 𝑡𝑎𝑛−1 0] = (2.0412)[0.6847 − 0] = 𝟏. 𝟑𝟗𝟕𝟔, correct to 4 decimal
places.
Exercise
3
1.Determine ∫ 4+𝑡 2 𝑑𝑡.
28
5
2.Determine ∫ 16+9𝜃2 𝑑𝜃.
1
3. Evaluate ∫0
1+𝑡 2
3
4.Evaluate ∫0
3
5
4+𝑥 2
𝑑𝑡.
𝑑𝑥.
Integration using 𝐬𝐢𝐧𝐡 𝜽 substitution
Examples
1.Determine ∫
1
√(𝑥 2 +𝑎2 )
𝑑𝑥
Let 𝑥 = 𝑎 sinh 𝜃, 𝑡ℎ𝑒𝑛
Hence ∫
1
√(𝑥 2 +𝑎2 )
𝑑𝑥
𝑑𝜃
𝑑𝑥 = ∫
= 𝑎 cosh 𝜃 𝑎𝑛𝑑 𝑑𝑥 = 𝑎 cosh 𝜃𝑑𝜃
1
√(𝑎2 𝑠𝑖𝑛ℎ2 𝜃+𝑎2 )
(𝑎 cosh 𝜃𝑑𝜃) = ∫
𝑎 cosh 𝜃
√𝑎2 𝑐𝑜𝑠ℎ2 𝜃
𝑑𝜃,
𝑠𝑖𝑛𝑐𝑒 𝑐𝑜𝑠ℎ2 𝜃 − 𝑠𝑖𝑛ℎ2 𝜃 = 1.
= ∫ 𝑑𝜃 = 𝜃 + 𝑐 = 𝒔𝒊𝒏𝒉−𝟏
𝒙
+ 𝒄 , 𝑠𝑖𝑛𝑐𝑒 𝑥 = 𝑎 sinh 𝜃.
𝒂
Inverse hyperbolic functions may be evaluated most conveniently when expressed in a
logarithmic form.
𝑥
Let 𝑦 = 𝑠𝑖𝑛ℎ−1 𝑎 , 𝑡ℎ𝑒𝑛
𝑥
𝑎
= sinh 𝑦.
From definition of hyperbolic functions, 𝑒 𝑦 = cosh 𝑦 + sinh 𝑦
But cosh 𝑦 = √𝑠𝑖𝑛ℎ2 𝑦 + 1.
𝑥
𝑥
Therefore 𝑒 𝑦 = √(𝑎)2 + 1 + 𝑎 =
√𝑥 2 +𝑎2
𝑎
Thus taking natural logarithms, 𝑦 = ln
Hence ∫
1
√(𝑥 2 +𝑎2 )
2
2.Evaluate ∫0
2
∫0
1
√𝑥 2 +4
𝑑𝑥 = ln {
1
√𝑥 2 +4
𝑥+√𝑥 2 +𝑎2
𝑎
𝑥
+𝑎=
𝑥+√𝑥2 +𝑎2
𝑎
𝑥+√𝑥 2 +𝑎2
𝑎
𝑥
𝑥+√𝑥 2 +𝑎2
𝑎
𝑎
or 𝑠𝑖𝑛ℎ−1 = ln
}+𝑐
𝑑𝑥, correct to 4 decimal places.
𝑥 2
𝑥+√𝑥2 +4 2
𝑑𝑥 = [𝑠𝑖𝑛ℎ−1 2] 𝑜𝑟 [ln { 2 }] , since a = 2.
0
0
29
.
2
Using the logarithmic form, ∫0
2
3.Evaluate ∫1
2
𝑥 2 √1+𝑥 2
1
√𝑥 2 +4
𝑑𝑥 = [ln(
2+√8
0+√4
2
2
) − ln(
)] = ln 2.4142 − ln 1 = 𝟎. 𝟖𝟖𝟏𝟒
𝑑𝑥, correct to 3 significant figures
Since the integral contains a term of the form √𝑎2 + 𝑥 2 , then let 𝑥 =
𝑎 sinh 𝜃, 𝑓𝑟𝑜𝑚 𝑤ℎ𝑖𝑐ℎ
Hence ∫
2
𝑥 2 √1+𝑥 2
𝑑𝑥
𝑑𝜃
𝑑𝑥 = ∫
= 𝑎 cosh 𝜃 𝑎𝑛𝑑 𝑑𝑥 = 𝑎 cosh 𝜃𝑑𝜃
2(cosh 𝜃𝑑𝜃)
𝑠𝑖𝑛ℎ2 𝜃√(1+𝑠𝑖𝑛ℎ2 𝜃)
= 2∫
𝑑𝜃
= 2 ∫ 𝑐𝑜𝑠𝑒𝑐ℎ2 𝜃 𝑑𝜃 = −2 coth 𝜃 + 𝑐
𝑠𝑖𝑛ℎ2 𝜃
coth 𝜃 =
2
Hence ∫1
cosh 𝜃𝑑𝜃
= 2 ∫ 𝑠𝑖𝑛ℎ2 𝜃 cosh 𝜃, 𝑠𝑖𝑛𝑐𝑒 𝑐𝑜𝑠ℎ2 𝜃 − 𝑠𝑖𝑛ℎ2 𝜃 = 1
cosh 𝜃 √(1 + 𝑠𝑖𝑛ℎ2 𝜃) √1 + 𝑥 2
=
=
sinh 𝜃
sinh 𝜃
𝑥
√1+𝑥 2 2
√5
√2
2
𝑑𝑥
=
−[2
coth
𝜃]
=
−2
[
] = −2 [ 2 − 1 ] = 0.592,
𝑥
𝑥 2 √1+𝑥 2
1
1
2
correct to 3 significant figures.
4.Find ∫ √(𝑥 2 + 𝑎2 ) 𝑑𝑥
Let 𝑥 = 𝑎 sinh 𝜃, 𝑡ℎ𝑒𝑛
𝑑𝑥
𝑑𝜃
= 𝑎 cosh 𝜃 𝑎𝑛𝑑 𝑑𝑥 = 𝑎 cosh 𝜃𝑑𝜃
Hence ∫ √(𝑥 2 + 𝑎2 ) 𝑑𝑥 = ∫ √(𝑎2 𝑠𝑖𝑛ℎ2 𝜃 + 𝑎2 )(𝑎 cosh 𝜃𝑑𝜃) =
∫ √𝑎2 (𝑠𝑖𝑛ℎ2 𝜃 + 1) (𝑎 cosh 𝜃𝑑𝜃)
= ∫ √𝑎2 𝑐𝑜𝑠ℎ2 𝜃 (𝑎 cosh 𝜃𝑑𝜃) = ∫(𝑎 cosh 𝜃) (𝑎 cosh 𝜃)𝑑𝜃 = 𝑎2 ∫ 𝑐𝑜𝑠ℎ2 𝜃 𝑑 𝜃
= 𝑎2 ∫ (
1 + cosh 2𝜃
𝑎2
sinh 2𝜃
𝑎2
) 𝑑𝜃 = (𝜃 +
) + 𝑐 = (𝜃 + sinh 𝜃 cosh 𝜃) + 𝑐,
2
2
2
2
Since sinh 2𝜃 = 2 sinh 𝜃 cosh 𝜃.
𝑥
𝑥
Since 𝑥 = asinh 𝜃, 𝑡ℎ𝑒𝑛 sinh 𝜃 = 𝑎 𝑎𝑛𝑑 𝜃 = 𝑠𝑖𝑛ℎ−1 𝑎
𝑥
Also since 𝑐𝑜𝑠ℎ2 𝜃 − 𝑠𝑖𝑛ℎ2 𝜃 = 1, then cosh 𝜃 = √(1 + 𝑠𝑖𝑛ℎ2 𝜃) = √[1 + (𝑎)2 ] = √(
=
√𝑎2 + 𝑥 2
𝑎
30
𝑎2 +𝑥 2
𝑎2
)
𝑎2
Hence ∫ √(𝑥 2 + 𝑎2 ) 𝑑𝑥 =
2
𝑥 √𝑎2 +𝑥2
𝑥
[𝑠𝑖𝑛ℎ−1 𝑎 + (𝑎)
𝑎
]+𝑐 =
𝒂𝟐
𝟐
𝒙
𝒙
𝒔𝒊𝒏𝒉−𝟏 𝒂 + 𝟐 √𝒂𝟐 + 𝒙𝟐 + 𝒄
Exercise
1.Find ∫
2
√(𝑥 2 +16)
𝑑𝑥,
2.Find∫ √(𝑥 2 + 9) 𝑑𝑥,
3
3.Evaluate ∫0
4
√(𝑡 2 +9)
𝑑𝑡,
1
4.Evaluate ∫0 √(16 + 9𝜃 2 ) 𝑑𝜃.
Integration using 𝐜𝐨𝐬𝐡 𝜽 substitution
1
1.Determine ∫ √𝑥 2
−𝑎2
𝑑𝑥.
Let 𝑥 = 𝑎 cosh 𝜃 𝑡ℎ𝑒𝑛
1
Hence ∫ √𝑥 2
𝑑𝑥
= asinh 𝜃 𝑎𝑛𝑑 𝑑𝑥 = asinh 𝜃𝑑𝜃
𝑑𝜃
1
−𝑎2
asinh 𝜃𝑑𝜃
asinh 𝜃𝑑𝜃
𝑑𝑥 = ∫ 2
(asinh 𝜃𝑑𝜃) = ∫ 2
= ∫ √𝑎2
,
𝑠𝑖𝑛ℎ2 𝜃
√(𝑎 𝑐𝑜𝑠ℎ2 𝜃−𝑎2 )
√𝑎 (𝑐𝑜𝑠ℎ2 𝜃−1)
𝑠𝑖𝑛𝑐𝑒 𝑐𝑜𝑠ℎ2 𝜃 − 𝑠𝑖𝑛ℎ2 𝜃 = 1.
=∫
asinh 𝜃𝑑𝜃
𝒙
= ∫ 𝑑𝜃 = 𝜃 + 𝑐 = 𝒄𝒐𝒔𝒉−𝟏 + 𝒄 , 𝑠𝑖𝑛𝑐𝑒 𝑥 = acosh 𝜃.
asinh 𝜃
𝒂
𝑥
It can be shown that 𝑐𝑜𝑠ℎ−1 𝑎 = ln {
1
Hence ∫ √𝑥 2
−𝑎2
𝑑𝑥 = 𝐥𝐧 {
2𝑥−3
2.Determine ∫ √𝑥 2
−9
𝑥+√(𝑥 2 −𝑎2 )
𝑎
}
𝒙+√(𝒙𝟐 −𝒂𝟐 )
𝒂
} + 𝒄 provides an alternative solution.
𝑑𝑥
∫
2𝑥 − 3
√𝑥 2 − 9
𝑑𝑥 = ∫
2𝑥
√𝑥 2 − 9
𝑑𝑥 − ∫
3
√𝑥 2 − 9
𝑑𝑥
The first integral is determined using the algebraic substitution 𝑢 = 𝑥 2 − 9, and the second
1
integral is of the form∫ √𝑥 2
2𝑥
Hence ∫ √𝑥 2
−9
−𝑎2
3
𝑑𝑥 − ∫ √𝑥 2
𝑑𝑥.
𝒙
−9
𝑑𝑥 = 𝟐√(𝒙𝟐 − 𝟗 − 𝟑𝒄𝒐𝒔𝒉−𝟏 𝟑 + 𝒄
3.Determine ∫ √𝑥 2 − 𝑎2 𝑑𝑥
31
Let 𝑥 = 𝑎 cosh 𝜃 𝑡ℎ𝑒𝑛
𝑑𝑥
= asinh 𝜃 𝑎𝑛𝑑 𝑑𝑥 = asinh 𝜃𝑑𝜃
𝑑𝜃
Hence∫ √𝑥 2 − 𝑎2 𝑑𝑥 = ∫ √𝑎2 𝑐𝑜𝑠ℎ2 𝜃 − 𝑎2 (asinh 𝜃𝑑𝜃) = ∫ √𝑎2 (𝑐𝑜𝑠ℎ2 𝜃 − 1) (asinh 𝜃𝑑𝜃)
= ∫ √𝑎2 𝑠𝑖𝑛ℎ2 𝜃 (asinh 𝜃𝑑𝜃) = 𝑎2 ∫ 𝑠𝑖𝑛ℎ2 𝜃𝑑𝜃 = 𝑎2 ∫(
cosh 2𝜃 − 1
) 𝑑𝜃,
2
since cosh 2𝜃 = 1 + 2𝑠𝑖𝑛ℎ2 𝜃.
=
𝑎2 sinh 2𝜃
𝑎2
[
− 𝜃] + 𝑐 = [sinh 𝜃 cosh 𝜃 − 𝜃] + 𝑐, 𝑠𝑖𝑛𝑐𝑒 sinh 2𝜃 = 2 sinh 𝜃 cosh 𝜃.
2
2
2
𝑥
𝑥
Since 𝑥 = acosh 𝜃 𝑡ℎ𝑒𝑛 cosh 𝜃 = 𝑎 𝑎𝑛𝑑 𝜃 = 𝑐𝑜𝑠ℎ−1 𝑎.
𝑥
Also, since 𝑐𝑜𝑠ℎ2 𝜃 − 𝑠𝑖𝑛ℎ2 𝜃 = 1, 𝑡ℎ𝑒𝑛 sinh 𝜃 = √(𝑐𝑜𝑠ℎ2 𝜃 − 1) = √(𝑎)2 − 1 =
Hence ∫ √𝑥 2 − 𝑎2 𝑑𝑥 =
𝑎2 √(𝑥 2 −𝑎2 ) 𝑥
[
2
𝑎
𝑥
𝒙
(𝑎) − 𝑐𝑜𝑠ℎ−1 𝑎]+c = 𝟐 √(𝒙𝟐 − 𝒂𝟐 ) −
𝒂𝟐
𝟐
√(𝑥 2 −𝑎2 )
𝑎
𝒙
𝒄𝒐𝒔𝒉−𝟏 𝒂 + 𝒄
3
4.Evaluate∫2 √𝑥 2 − 4 𝑑𝑥.
3
𝑥
4
𝑥 3
From above example, ∫2 √𝑥 2 − 4 𝑑𝑥 = [2 √(𝑥 2 − 4) − 2 𝑐𝑜𝑠ℎ−1 2] , 𝑤ℎ𝑒𝑛 𝑎 = 2.
2
3
3
= ( √5 − 2𝑐𝑜𝑠ℎ−1 ) − (0 − 2𝑐𝑜𝑠ℎ−1 1)
5
2
𝑥
Since 𝑐𝑜𝑠ℎ−1 𝑎 = ln {
𝑥+√(𝑥 2 −𝑎2 )
𝑎
3
} , 𝑡ℎ𝑒𝑛 𝑐𝑜𝑠ℎ−1 2 = ln {
3+√(32 −22 )
2
} = ln 2.6180 = 0.9624
Similarly, 𝑐𝑜𝑠ℎ−1 1 = 0.
3
Hence ∫ √𝑥 2 − 𝑎2 𝑑𝑥 = [2 √5 − 2(0.9624)] − (0) = 𝟏. 𝟒𝟐𝟗, correct to 4 significant figures
Exercise
1
1.Find ∫ √𝑡 2
2.Find∫
−16
𝑑𝑡,
3
√(4𝑥 2 −9)
2
3.Evaluate∫1
𝑑𝑥,
2
√(𝑥 2 −1)
𝑑𝑥,
3
4.Evaluate ∫2 √(𝑡 2 − 4) 𝑑𝑡.
32
𝜽
The 𝒕 = 𝐭𝐚𝐧 𝟐 substitution
1
Integrals of the form ∫ acos 𝜃+𝑏 sin 𝜃+𝑐 𝑑𝜃, where a, b and c are constants, may be determined by
𝜃
using the substitution 𝑡 = tan 2.
. The reasoning is as explained below.
𝜃
If angle A in the right-angled triangle ABC shown in the figure below is made equal to 2 then,
𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒
𝜃
since tangent = 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡, if BC = t and AB = 1, then tan 2 = t.
C
√1 + 𝑡 2
𝑡
𝜃
2
A
1
B
By Pythagoras’ theorem, AC = √1 + 𝑡 2
Therefore,
sin
𝜃
𝑡
𝜃
1
=
𝑎𝑛𝑑 cos =
2 √1 + 𝑡 2
2 √1 + 𝑡 2
𝜃
𝜃
Since sin 2x = 2 sin x cos x(from double angle formulae), then sin 𝜃 = 2 sin 2 cos 2
= 2(
𝑡
√1 + 𝑡 2
)(
1
√1 + 𝑡 2
𝟐𝒕
i.e. 𝐬𝐢𝐧 𝜽 = 𝟏+𝒕𝟐 …(1)
33
)=
2𝑡
1 + 𝑡2
𝜃
𝜃
1
𝑡
Since cos 2x = 𝑐𝑜𝑠 2 𝑥 − 𝑠𝑖𝑛2 𝑥, then cos 𝜃 = 𝑐𝑜𝑠 2 2 − 𝑠𝑖𝑛2 2 = (√1+𝑡 2 )2 − (√1+𝑡 2 )2
𝟏−𝒕𝟐
i.e. 𝐜𝐨𝐬 𝜽 = 𝟏+𝒕𝟐 …(2)
𝜃
𝑑𝑡
1
𝜃
1
𝜃
Also, since t = tan 2 , 𝑑𝜃 = 2 𝑠𝑒𝑐 2 2 = 2 (1 + 𝑡𝑎𝑛2 2) from trigonometric identities,
𝑑𝑡
1
𝟐𝒅𝒕
i.e. 𝑑𝜃 = 2 (1 + 𝑡 2 ) from which 𝒅𝜽 = 𝟏+𝒕𝟐 …(3)
1
Equations (1), (2) and (3) are used to determine integrals of the form ∫ acos 𝜃+𝑏 sin 𝜃+𝑐 𝑑𝜃
where a, b or c may be zero.
Examples
1
1.Determine ∫ sin 𝜃 𝑑𝜃.
𝜃
2𝑡
2𝑑𝑡
If 𝑡 = tan 2, then sin 𝜃 = 1+𝑡 2 and 𝑑𝜃 = 1+𝑡 2 from equations (1) and (3).
1
Thus ∫ sin 𝜃 𝑑𝜃 = ∫
1
2𝑑𝑡
1
(1+𝑡 2 ) = ∫ 𝑡 𝑑𝑡 = ln 𝑡 + 𝑐
2𝑡
1+𝑡2
1
𝜽
Hence ∫ sin 𝜃 𝑑𝜃 = 𝐥𝐧(𝐭𝐚𝐧 𝟐) + 𝒄
𝑑𝑥
2.Determine ∫ cos 𝑥.
1−𝑡 2
𝑥
2𝑑𝑡
If 𝑡 = tan 2 then cos 𝑥 = 1+𝑡 2 and 𝑑𝑥 = 1+𝑡 2 from equations (2) and (3).
𝑑𝑥
1
2𝑑𝑡
2
Thus ∫ cos 𝑥 = ∫ 1−𝑡2 (1+𝑡 2 ) = ∫ 1−𝑡 2 𝑑𝑡.
1+𝑡2
2
1−𝑡 2
Let
may be resolved into partial fractions.
2
1−𝑡 2
2
≡ (1−𝑡)(1+𝑡) ≡
𝐴
1−𝑡
+
𝐵
1+𝑡
≡
𝐴(1+𝑡)+𝐵(1−𝑡)
(1−𝑡)(1+𝑡)
Hence 2 ≡ 𝐴(1 + 𝑡) + 𝐵(1 − 𝑡).
When t = 1, 2 = 2A, from which,A = 1
When t = −1, 2 = 2B, from which,B = 1
2
1
1
(1+𝑡)
Hence ∫ 1−𝑡 2 𝑑𝑡 = ∫ {1−𝑡 + 1+𝑡} 𝑑𝑡 = − ln(1 − 𝑡) + ln(1 + 𝑡) + 𝑐 = ln {(1−𝑡)} + 𝑐
34
𝒅𝒙
Thus ∫ 𝐜𝐨𝐬 𝒙 = 𝐥𝐧 {
𝒙
𝟐
𝒙
𝟏 −𝐭𝐚𝐧
𝟐
𝟏+𝐭𝐚𝐧
} + 𝒄.
𝜋
Note that since tan 4 = 1, the above result may be written as:
𝜋
𝑥
tan 4 + tan 2
𝑑𝑥
𝝅 𝒙
∫
= ln {
}
+
𝑐
=
𝐥𝐧
{𝐭𝐚𝐧(
+ )} + 𝒄
𝜋
𝑥
cos 𝑥
𝟒 𝟐
1 − tan 4 tan 2
𝑑𝑥
3.Determine ∫ 1+cos 𝑥.
1−𝑡 2
𝑥
2𝑑𝑡
If 𝑡 = tan 2 then cos 𝑥 = 1+𝑡 2 and 𝑑𝑥 = 1+𝑡 2 from equations (2) and (3).
𝑑𝑥
Thus∫ 1+cos 𝑥 = ∫
1
2𝑑𝑡
1
2𝑑𝑡
(1+𝑡 2 ) = ∫ (1+𝑡2 )+(1−𝑡2 ) (1+𝑡 2 ) = ∫ 𝑑𝑡
1−𝑡2
1+ 2
1+𝑡
1+𝑡2
𝑑𝑥
𝒙
Hence ∫ 1+cos 𝑥 = 𝑡 + 𝑐 = 𝐭𝐚𝐧 𝟐 + 𝒄
𝑑𝜃
4.Determine∫ 5+4 cos 𝜃 .
1−𝑡 2
𝜃
2𝑑𝑡
If 𝑡 = tan 2, then cos 𝜃 = 1+𝑡 2 and 𝑑𝜃 = 1+𝑡 2 from equations (1) and (3).
2𝑑𝑡
1+𝑡2
1−𝑡2
5+4( 2 )
1+𝑡
𝑑𝜃
Thus ∫ 5+4 cos 𝜃 = ∫
𝒅𝜽
𝟐
=∫
𝟏
2𝑑𝑡
1+𝑡2
5(1+𝑡2 )+4(1−𝑡2 )
(1+𝑡2 )
𝑑𝑡
𝑑𝑡
1
𝑡
= 2 ∫ 𝑡 2 +9 = 2 ∫ 𝑡 2 +32 = 2(3 𝑡𝑎𝑛−1 3) + 𝑐
𝜽
Hence ∫ 𝟓+𝟒 𝐜𝐨𝐬 𝜽 = 𝟑 𝒕𝒂𝒏−𝟏 (𝟑 𝐭𝐚𝐧 𝟐) + 𝒄
𝑑𝑥
5.Determine ∫ sin 𝑥+cos 𝑥 .
𝑥
1−𝑡 2
2𝑡
2𝑑𝑡
If 𝑡 = tan 2 then sin 𝑥 = 1+𝑡 2, , cos 𝑥 = 1+𝑡 2 and 𝑑𝑥 = 1+𝑡 2 from equations (1), (2) and (3).
Thus ∫
𝑑𝑥
sin 𝑥+cos 𝑥
=∫
2𝑑𝑡
1+𝑡2
2𝑡
1−𝑡2
+
1+𝑡2 1+𝑡2
=∫
𝒅𝒙
i.e. ∫ 𝐬𝐢𝐧 𝒙+𝐜𝐨𝐬 𝒙 =
𝟏
√𝟐
=∫
2𝑑𝑡
1+𝑡2
2𝑡+1−𝑡2
1+𝑡2
=∫
2𝑑𝑡
1+2𝑡−𝑡 2
=∫
−2𝑑𝑡
𝑡 2 −2𝑡−1
=∫
−2𝑑𝑡
(𝑡−1)2 −2
2𝑑𝑡
1
√2 + (𝑡 − 1)
= 2[
ln {
}] + 𝑐
(√2)2 − (𝑡 − 1)2
2√2
√2 − (𝑡 − 1)
𝒙
𝐥𝐧 {
√𝟐−𝟏+𝐭𝐚𝐧𝟐
𝒙
√𝟐+𝟏−𝐭𝐚𝐧𝟐
} + 𝒄 (See problem 9 under Integration using partial fractions)
35
𝑑𝑥
6.Determine ∫ 7−3 sin 𝑥+6 cos 𝑥.
From equations (1), (2) and (3),
𝑑𝑥
∫
=∫
7 − 3 sin 𝑥 + 6 cos 𝑥
=∫
7+
7𝑡 2
2𝑑𝑡
2𝑑𝑡
1 + 𝑡2
1
+ 𝑡2
=∫
2
2𝑡
1−𝑡
7(1 + 𝑡 2 ) − 3(2𝑡) + 6(1 − 𝑡 2 )
7 − 3(
) + 6(
)
2
2
1+𝑡
1+𝑡
1 + 𝑡2
2𝑑𝑡
2𝑑𝑡
2𝑑𝑡
1
𝑡−3
=∫ 2
=∫
= 2 [ 𝑡𝑎𝑛−1 (
)] + 𝑐
2
2
2
− 6𝑡 + 6 − 6𝑡
𝑡 − 6𝑡 + 13
(𝑡 − 3) + 2
2
2
𝒙
𝟐
𝐭𝐚𝐧 −𝟑
𝑑𝑥
Hence ∫ 7−3 sin 𝑥+6 cos 𝑥 = 𝒕𝒂𝒏−𝟏 (
)+𝒄
𝟐
𝑑𝜃
7.Determine ∫ 4 cos 𝜃+3 sin 𝜃.
From equations (1) to (3),
𝑑𝜃
∫
=∫
4 cos 𝜃 + 3 sin 𝜃
2𝑑𝑡
1 + 𝑡2
4(
𝑡2
1−
2𝑡
) + 3(
)
2
1+𝑡
1 + 𝑡2
=∫
2𝑑𝑡
𝑑𝑡
=
∫
4 − 4𝑡 2 + 6𝑡
2 + 3𝑡 − 2𝑡 2
1
𝑑𝑡
1
𝑑𝑡
1
𝑑𝑡
=− ∫
=− ∫
= ∫
2 𝑡2 − 3 𝑡 − 1
2 (𝑡 − 3)2 − 25 2 (5)2 − (𝑡 − 3)2
2
4
16
4
4
5
3
1
+ (𝑡 − 4)
+𝑡
1 1
1
= [
ln {4
} + 𝑐] = ln {2
}+𝑐
5
3
2 2(5)
5
2−𝑡
−
(𝑡
−
)
4
4
4
(See problem 9 under Integration using partial fractions)
𝑑𝜃
𝟏
𝟏
𝜽
𝟐
𝜽
𝟐−𝐭𝐚𝐧
𝟐
Hence ∫ 4 cos 𝜃+3 sin 𝜃 = 𝟓 𝐥𝐧 { 𝟐
+𝐭𝐚𝐧
𝟏
} + 𝒄 or 𝟓 𝐥𝐧 {
𝜽
𝟐
𝜽
𝟒−𝟐 𝐭𝐚𝐧
𝟐
𝟏+𝟐𝐭𝐚𝐧
}+𝒄
Partial Fractions
By algebraic addition,
(𝑥 + 1) + 3(𝑥 − 2)
1
3
4𝑥 − 5
+
=
= 2
(𝑥 − 2)(𝑥 + 1)
𝑥−2 𝑥+1
𝑥 −𝑥−2
4𝑥−5
1
3
The reverse process of moving from 𝑥 2 −𝑥−2 to 𝑥−2 + 𝑥+1 is called resolving into partial
fractions.
36
In order to resolve an algebraic expression into partial fractions:
(i) the denominator must factorize (in the above example, 𝑥 2 − 𝑥 − 2 factorizes as (x−2)(x+1)),
and
(ii) the numerator must be at least one degree less than the denominator (in the above example
(4x−5) is of degree 1 since the highest powered x term is 𝑥1 and (𝑥 2 − 𝑥 − 2) is of degree 2).
When the degree of the numerator is equal to or higher than the degree of the denominator, the
numerator must be divided by the denominator until the remainder is of less degree than the
denominator(see Problems (c) and (d)).
There are basically three types of partial fraction and the form of partial fraction used is
summarized in the following table , where f(x) is assumed to be of less degree than the relevant
denominator and A, B and Care constants to be determined.
(In the latter type in the table , 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 is a quadratic expression which does not factorize
without containing surds or imaginary terms.)
Resolving an algebraic expression into partial fractions is used as a preliminary to integrating
certain functions.
Type
1
Denominator containing
Linear factors
2
Repeated linear factors
3
Quadratic factors
Expression
𝑓(𝑥)
(𝑥 + 𝑎)(𝑥 − 𝑏)(𝑥 + 𝑐)
𝑓(𝑥)
(𝑥 + 𝑎)3
𝑓(𝑥)
2
(𝑎𝑥 + 𝑏𝑥 + 𝑐)(𝑥 + 𝑑)
Form of Partial Fraction
𝐴
𝐵
𝐶
+
+
(𝑥 + 𝑎) (𝑥 − 𝑏) (𝑥 + 𝑐)
𝐴
𝐵
𝐶
+
+
2
(𝑥 + 𝑎) (𝑥 + 𝑎)
(𝑥 + 𝑎)3
𝐴𝑥 + 𝐵
𝐶
+
2
(𝑎𝑥 + 𝑏𝑥 + 𝑐) (𝑥 + 𝑑)
Examples
(a)Resolve
11−3𝑥
𝑥 2 +2𝑥−3
into partial fractions.
The denominator factorizes as (x−1) (x+3) and the numerator is of less degree than the
denominator. Thus
Let
11−3𝑥
𝑥 2 +2𝑥−3
11−3𝑥
𝑥 2 +2𝑥−3
11−3𝑥
may be resolved into partial fractions.
𝐴
≡ (𝑥−1)(𝑥+3) ≡ (𝑥−1) +
𝐵
(𝑥+3)
Where A and Bare constants to be determined,
37
11−3𝑥
𝐴(𝑥+3)+𝐵(𝑥−1)
(𝑥−1)(𝑥+3)
i.e. 𝑥 2 +2𝑥−3 ≡
by algebraic addition.
Since the denominators are the same on each side of the identity then the numerators are equal to
each other.
Thus, 11−3x ≡ A(x+3)+B(x−1)
To determine constants A and B, values of x are chosen to make the term in A or B equal to zero.
When x=1, then 11−3(1) ≡ A(1+3)+B(0)
i.e. 8 = 4A
i.e. A = 2
When x = −3, then 11−3(−3) ≡ A(0)+B(−3−1)
i.e. 20 = −4B
i.e. B = −5
11−3𝑥
2
−5
2
5
Thus, 𝑥 2 +2𝑥−3 = (𝑥−1) + (𝑥+3) = (𝑥−1) − (𝑥+3)
Check: The right hand side simplifies to the result on the left by simple subtraction of fractions.
2𝑥 2 −9𝑥−35
(b) Convert (𝑥+1)(𝑥−2)(𝑥+3) into the sum of three partial fractions.
2𝑥 2 −9𝑥−35
𝐴
𝐵
𝐶
Let (𝑥+1)(𝑥−2)(𝑥+3) ≡ (𝑥+1) + (𝑥−2) + (𝑥+3) ≡
𝐴(𝑥−2)(𝑥+3)+𝐵(𝑥+1)(𝑥+3)+𝐶(𝑥+1)(𝑥−2)
(𝑥+1)(𝑥−2)(𝑥+3)
by algebraic addition.
Equating the numerators gives:
2𝑥 2 − 9x – 35 ≡ A(x−2)(x+3) + B(x+1)(x+3) + C(x+1)(x−2)
Let x = −1. Then
2(−1)2 − 9(−1) – 35 ≡ A(−3)(2) + B(0)(2) + C(0)(−3)
i.e. −24 = −6A
i.e. A = 4
Let x = 2. Then
2(2)2 − 9(2) – 35 ≡ A(0)(5) + B(3)(5) + C(3)(0)
38
i.e. −45 = 15B
i.e. B = −3
Let x = −3. Then
2(−3)2 − 9(−3) – 35 ≡ A(−5)(0) + B(−2)(0) + C(– 2)(–5)
i.e. 10 = 10 C
i.e. C = 1
2𝑥 2 −9𝑥−35
4
3
1
Thus (𝑥+1)(𝑥−2)(𝑥+3) ≡ (𝑥+1) − (𝑥−2) + (𝑥+3)
𝑥 2 +1
(c)Resolve 𝑥 2 −3𝑥+2 into partial fractions.
The denominator is of the same degree as the numerator. Thus dividing out gives:
1
2 +
̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅
𝑥 2 − 3𝑥 + 2)𝑥
1
𝑥 2 − 3𝑥 + 2
3𝑥 − 1
𝑥 2 +1
3𝑥−1
3𝑥−1
Hence 𝑥 2 −3𝑥+2 ≡ 1 + 𝑥 2 −3𝑥+2 ≡ 1 + (𝑥−1)(𝑥−2).
3𝑥−1
𝐴
𝐵
Let (𝑥−1)(𝑥−2) ≡ (𝑥−1) + (𝑥−2) ≡
𝐴(𝑥−2)+𝐵(𝑥−1)
.
(𝑥−1)(𝑥−2)
Equating numerators gives:
3x−1 ≡ A(x−2) + B(x−1)
Let x=1.Then 2 = −A
i.e. A = −2
Let x = 2.Then 5 = B
3𝑥−1
−2
5
Hence (𝑥−1)(𝑥−2) ≡ (𝑥−1) + (𝑥−2).
𝑥 2 +1
2
5
Thus 𝑥 2 −3𝑥+2 ≡ 1 − (𝑥−1) + (𝑥−2).
39
(d)Express
𝑥 3 −2𝑥 2 −4𝑥−4
𝑥 2 +𝑥−2
in partial fractions.
The numerator is of higher degree than the denominator. Thus dividing out gives:
𝑥−3
3 − 2𝑥 2 − 4𝑥 − 4
̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅
𝑥 2 + 𝑥 − 2)𝑥
𝑥 3 + 𝑥 2 − 2𝑥
−3𝑥 2 − 2𝑥 − 4
−3𝑥 2 − 3𝑥 + 6
𝑥 − 10
𝑥 3 −2𝑥 2 −4𝑥−4
Thus
𝑥 2 +𝑥−2
𝑥−10
𝑥−10
𝑥−10
≡ 𝑥 − 3 + 𝑥 2 +𝑥−2 ≡ 𝑥 − 3 + (𝑥+2)(𝑥−1)
𝐴
𝐵
Let (𝑥+2)(𝑥−1) ≡ (𝑥+2) + (𝑥−1) ≡
𝐴(𝑥−1)+𝐵(𝑥+2)
(𝑥+2)(𝑥−1)
Equating the numerators gives:
x − 10 ≡ A(x−1) + B(x+2)
Let x = −2.Then −12 = −3A
i.e. A = 4
Let x = 1.Then −9 = 3B
i.e. B = −3
𝑥−10
4
3
Hence (𝑥+2)(𝑥−1) ≡ (𝑥+2) − (𝑥−1)
Thus
𝑥 3 −2𝑥 2 −4𝑥−4
𝑥 2 +𝑥−2
4
3
≡ 𝑥 − 3 + (𝑥+2) − (𝑥−1)
2𝑥+3
(e)Resolve (𝑥−2)2 into partial fractions.
The denominator contains a repeated linear factor, (𝑥 − 2)2 .
2𝑥+3
𝐴
𝐵
Let (𝑥−2)2 ≡ (𝑥−2) + (𝑥−2)2 ≡
𝐴(𝑥−2)+𝐵
(𝑥−2)2
Equating the numerators gives:
40
2x + 3 ≡ A(x−2) + B
Let x = 2. Then 7 = A(0) + B
i.e. B = 7
2x + 3 ≡ A(x−2) + B ≡ Ax−2A + B
Since an identity is true for all values of the unknown, the coefficients of similar terms may be
equated. Hence, equating the coefficients of x gives:2 = A.
[Also, as a check, equating the constant terms gives: 3 = −2A + B
When A= 2 and B = 7,
R.H.S. = −2(2) + 7 = 3 = L.H.S.]
2𝑥+3
2
7
Hence (𝑥−2)2 ≡ (𝑥−2) + (𝑥−2)2
5𝑥 2 −2𝑥−19
(f)Express (𝑥+3)(𝑥−1)2 as the sum of three partial fractions.
The denominator is a combination of a linear factor and a repeated linear factor.
5𝑥 2 −2𝑥−19
Let (𝑥+3)(𝑥−1)2 ≡
𝐴
(𝑥+3)
+
𝐵
(𝑥−1)
+
𝐶
(𝑥−1)2
≡
𝐴(𝑥−1)2 +𝐵(𝑥+3)(𝑥−1)+𝐶(𝑥+3)
(𝑥+3)(𝑥−1)2
by algebraic addition.
Equating the numerators gives:
5𝑥 2 −2x −19 ≡ A(𝑥 − 1)2 + B(x+3)(x−1) + C(x+3)
(1)
Let x = −3. Then
5(−3)2 −2(−3) – 19 ≡ A(−4)2 + B(0)(−4) + C(0)
i.e. 32 = 16A, or A=2
Let x = 1. Then
5(1)2 − 2(1) – 19 ≡ A(0)2 + B(4)(0) + C(4)
i.e. −16 = 4C or C = −4
Without expanding the RHS of equation (1) it can be seen that equating the coefficients of 𝑥 2
gives:5 = A + B, and since A = 2, B = 3.
41
5𝑥 2 −2𝑥−19
2
3
4
Hence (𝑥+3)(𝑥−1)2 ≡ (𝑥+3) + (𝑥−1) − (𝑥−1)2.
Practice
Resolve
3𝑥 2 +16𝑥+15
(𝑥+3)3
3
into partial fractions.
2
6
Ans (𝑥+3) − (𝑥+3)2 − (𝑥+3)3
7𝑥 2 +5𝑥+13
(g) Express (𝑥 2 +2)(𝑥+1) in partial fractions.
The denominator is a combination of a quadratic factor, (𝑥 2 + 2), which does not factorize
without introducing imaginary surd terms, and a linear factor, (x+1). Let,
(𝐴𝑥 + 𝐵)(𝑥 + 1) + 𝐶(𝑥 2 + 2)
7𝑥 2 + 5𝑥 + 13
𝐴𝑥 + 𝐵
𝐶
≡
+
≡
(𝑥 2 + 2)(𝑥 + 1) (𝑥 2 + 2) (𝑥 + 1)
(𝑥 2 + 2)(𝑥 + 1)
Equating numerators gives:
7𝑥 2 + 5𝑥 + 13 ≡ (Ax+B)(x+1) + C(𝑥 2 + 2)
(1)
Let x = −1.Then
7(−1)2 + 5(−1) + 13 ≡ (Ax+B)(0) + C(1+2)
i.e. 15 = 3C or C = 5.
Identity (1) may be expanded as:
7𝑥 2 + 5𝑥 + 13 ≡ A𝑥 2 + Ax + Bx + B + C𝑥 2 + 2C
Equating the coefficients of 𝑥 2 terms gives:
7 = A + C, and since C = 5, A = 2.
Equating the coefficients of x terms gives:
5 = A + B, and since A = 2, B = 3.
7𝑥 2 +5𝑥+13
2𝑥+3
5
Hence (𝑥 2 +2)(𝑥+1) ≡ (𝑥 2 +2) + (𝑥+1).
Practice
Resolve
3+6𝑥+4𝑥 2 −2𝑥 3
𝑥 2 (𝑥 2 +3)
into partial fractions.
42
2
1
3−4𝑥
Ans 𝑥 + 𝑥 2 + 𝑥 2 +3
Integration using partial fractions
The process of expressing a fraction in terms of simpler fractions—called partial fractions—is
referred to as resolving the given fraction into partial fractions.
Certain functions have to be resolved into partial fractions before they can be integrated as
demonstrated in the following worked problems.
11−3𝑥
1.Determine ∫ 𝑥 2 +2𝑥−3 𝑑𝑥.
11−3𝑥
11−3𝑥
2
5
Now 𝑥 2 +2𝑥−3 = (𝑥−1)(𝑥+3) ≡ (𝑥−1) − (𝑥+3).
11−3𝑥
2
5
2𝑑𝑥
5𝑑𝑥
Hence∫ 𝑥 2 +2𝑥−3 𝑑𝑥 = ∫ {(𝑥−1) − (𝑥+3)} 𝑑𝑥 = ∫ (𝑥−1) − ∫ (𝑥+3) = 𝟐 𝐥𝐧(𝒙 − 𝟏) − 𝟓 𝐥𝐧(𝒙 + 𝟑) + 𝒄
(𝒙−𝟏)𝟐
Or 𝐥𝐧 {(𝒙+𝟑)𝟓 } + 𝒄 by use of the laws of logarithms.
2𝑥 2 −9𝑥−35
2.Find ∫ (𝑥+1)(𝑥−2)(𝑥+3) 𝑑𝑥.
2𝑥 2 − 9𝑥 − 35
4
3
1
≡
−
+
(𝑥 + 1)(𝑥 − 2)(𝑥 + 3) (𝑥 + 1) (𝑥 − 2) (𝑥 + 3)
2𝑥 2 −9𝑥−35
4
3
1
4𝑑𝑥
3𝑑𝑥
1𝑑𝑥
Hence ∫ (𝑥+1)(𝑥−2)(𝑥+3) 𝑑𝑥 = ∫ {(𝑥+1) − (𝑥−2) + (𝑥+3)} 𝑑𝑥 = ∫ (𝑥+1) − ∫ (𝑥−2) + ∫ (𝑥+3)
(𝒙 + 𝟏)𝟒 (𝒙 + 𝟑)
= 𝟒 𝐥𝐧(𝒙 + 𝟏) − 𝟑 𝐥𝐧(𝒙 − 𝟐) + 𝐥𝐧(𝒙 + 𝟑) + 𝒄 𝒐𝒓 𝐥𝐧 {
}+𝒄
(𝒙 − 𝟐)𝟑
𝑥 2 +1
3.Determine ∫ 𝑥 2 −3𝑥+2 𝑑𝑥.
By dividing out (since the numerator and denominator are of the same degree) and resolving into
𝑥 2 +1
2
5
partial fractions it was shown in (c) that 𝑥 2 −3𝑥+2 ≡ 1 − (𝑥−1) + (𝑥−2).
𝑥 2 +1
2
5
Hence ∫ 𝑥 2 −3𝑥+2 𝑑𝑥 = ∫ [1 − (𝑥−1) + (𝑥−2)] 𝑑𝑥 = 𝒙 − 𝟐 𝐥𝐧(𝒙 − 𝟏) + 𝟓 𝐥𝐧(𝒙 − 𝟐) + 𝒄 or
(𝒙 − 𝟐)𝟓
= 𝐱 + 𝐥𝐧 {
}+𝒄
(𝒙 − 𝟏)𝟐
3 𝑥 3 −2𝑥 2 −4𝑥−4
4.Evaluate ∫2
𝑥 2 +𝑥−2
𝑑𝑥, correct to 4 significant figures.
43
By dividing out and resolving into partial fractions it was shown in (d) that
𝑥 3 − 2𝑥 2 − 4𝑥 − 4
4
3
≡𝑥−3+
−
2
𝑥 +𝑥−2
(𝑥 + 2) (𝑥 − 1)
3 𝑥 3 −2𝑥 2 −4𝑥−4
Hence ∫2
𝑥 2 +𝑥−2
3
4
3
𝑑𝑥 = ∫2 {𝑥 − 3 + (𝑥+2) − (𝑥−1)} 𝑑𝑥
𝑥2
3
= [ − 3𝑥 + 4 ln(𝑥 + 2) − 3 ln(𝑥 − 1)]
2
2
9
= ( − 9 + 4 ln 5 − 3 ln 2) − (2 − 6 + 4 ln 4 − 3 ln 1) = −𝟏. 𝟔𝟖𝟕,
2
correct to 4 significant figures.
2𝑥+3
5.Determine ∫ (𝑥−2)2 𝑑𝑥.
2𝑥+3
2
7
It was shown in (e) that (𝑥−2)2 ≡ (𝑥−2) + (𝑥−2)2 .
2𝑥+3
2
7
𝟕
Thus ∫ (𝑥−2)2 𝑑𝑥 = ∫ {(𝑥−2) + (𝑥−2)2 } 𝑑𝑥 = 𝟐 𝐥𝐧(𝒙 − 𝟐) − (𝒙−𝟐) + 𝒄
5𝑥 2 −2𝑥−19
6.Find ∫ (𝑥+3)(𝑥−1)2 𝑑𝑥.
5𝑥 2 −2𝑥−19
2
3
4
It was shown in (f) that (𝑥+3)(𝑥−1)2 ≡ (𝑥+3) + (𝑥−1) − (𝑥−1)2.
5𝑥 2 −2𝑥−19
2
3
4
4
Hence ∫ (𝑥+3)(𝑥−1)2 𝑑𝑥 = ∫ {(𝑥+3) + (𝑥−1) − (𝑥−1)2 } 𝑑𝑥 = 2 ln(𝑥 + 3) + 3 ln(𝑥 − 1) + (𝑥−1) + 𝑐
𝟒
or 𝐥𝐧{(𝒙 + 𝟑)𝟐 (𝒙 − 𝟏)𝟑 } + (𝒙−𝟏) + 𝒄.
Practice
𝑥 2 +1
Find ∫ 𝑥 3 +2𝑥 2 +𝑥 𝑑𝑥.
1 3𝑥 2 +16𝑥+15
7.Evaluate ∫−2
(𝑥+3)3
𝑑𝑥, correct to 4 significant figures.
From the practice question above,
1 3𝑥 2 +16𝑥+15
Hence ∫−2
(𝑥+3)3
1
3𝑥 2 +16𝑥+15
(𝑥+3)3
3
≡
3
(𝑥+3)
2
6
2
6
− (𝑥+3)2 − (𝑥+3)3.
2
3
𝑑𝑥 = ∫−2 {(𝑥+3) − (𝑥+3)2 − (𝑥+3)3 } 𝑑𝑥 = [3 ln(𝑥 + 3) + (𝑥+3) + (𝑥+3)2 ]
44
1
−2
2
3
2
3
= (3 ln 4 + 4 + 16) − (3 ln 1 + 1 + 1) = −𝟎. 𝟏𝟓𝟑𝟔, correct to 4 significant figures.
8.Find ∫
3+6𝑥+4𝑥 2 −2𝑥 3
𝑥 2 (𝑥 2 +3)
𝑑𝑥.
3+6𝑥+4𝑥 2 −2𝑥 3
From the practice question above,
Thus∫
3+6𝑥+4𝑥 2 −2𝑥 3
𝑥 2 (𝑥 2 +3)
2
𝑥 2 (𝑥 2 +3)
1
≡
2
𝑥
1
3−4𝑥
+ 𝑥 2 + 𝑥 2 +3.
3−4𝑥
2
1
3
4𝑥
𝑑𝑥 = ∫ (𝑥 + 𝑥 2 + 𝑥 2 +3) 𝑑𝑥 = ∫ {𝑥 + 𝑥 2 + 𝑥 2 +3 − 𝑥 2 +3} 𝑑𝑥
∫
𝑥2
3
1
3
𝑥
𝑑𝑥 = 3 ∫
𝑑𝑥 =
𝑡𝑎𝑛−1
+3
𝑥 2 + (√3)2
√3
√3
4𝑥
∫ 𝑥 2 +3 𝑑𝑥 is determined using the algebraic substitution 𝑢 = 𝑥 2 + 3.
2
1
3
4𝑥
1
Hence ∫ {𝑥 + 𝑥 2 + 𝑥 2 +3 − 𝑥 2 +3} 𝑑𝑥 = 2 ln 𝑥 − 𝑥 +
= 𝐥𝐧(
3
√3
𝑡𝑎𝑛−1
𝑥
√3
− 2 ln(𝑥 2 + 3) + 𝑐
𝒙
𝟏
𝒙
𝟐
−𝟏
)
−
+
+𝒄
√𝟑𝒕𝒂𝒏
𝒙𝟐 + 𝟑
𝒙
√𝟑
1
9.Determine ∫ 𝑎2 −𝑥 2 𝑑𝑥.
Using partial fractions, let
𝑎2
1
1
𝐴
𝐵
𝐴(𝑎 + 𝑥) + 𝐵(𝑎 − 𝑥)
≡
≡
+
≡
2
(𝑎 − 𝑥)(𝑎 + 𝑥) (𝑎 − 𝑥) (𝑎 + 𝑥)
(𝑎 − 𝑥)(𝑎 + 𝑥)
−𝑥
Then 1 ≡ A(a + x) + B(a − x)
1
1
Let x = a then A = 2𝑎. Let x = −a then B = 2𝑎
1
1
1
1
1
Hence ∫ 𝑎2 −𝑥 2 𝑑𝑥 = ∫ 2𝑎 [(𝑎−𝑥) + (𝑎+𝑥)] 𝑑𝑥 = 2𝑎 [− ln(𝑎 − 𝑥) + ln(𝑎 + 𝑥)] + 𝑐
=
𝟏
𝒂+𝒙
𝐥𝐧(
)+𝒄
𝟐𝒂
𝒂−𝒙
Exercise
1
Determine ∫ 𝑥 2 −𝑎2 𝑑𝑥.
45
Integration by parts
From the product rule of differentiation:
𝑑
𝑑𝑢
𝑑𝑣
(𝑢𝑣) = 𝑣
+𝑢 ,
𝑑𝑥
𝑑𝑥
𝑑𝑥
where u and v are both functions of x.
𝑑𝑣
𝑑
𝑑𝑢
Rearranging gives: 𝑢 𝑑𝑥 = 𝑑𝑥 (𝑢𝑣) − 𝑣 𝑑𝑥 .
Integrating both sides with respect to x gives:
∫𝑢
𝑑𝑣
𝑑𝑣
𝑑
𝑑𝑢
𝑑𝑥 = ∫ (𝑢𝑣) 𝑑𝑥 − ∫ 𝑣
𝑑𝑥
𝑑𝑥
𝑑𝑥
𝑑𝑥
𝑑𝑢
i.e. ∫ 𝑢 𝑑𝑥 𝑑𝑥 = 𝑢𝑣 − ∫ 𝑣 𝑑𝑥 𝑑𝑥 𝑜𝑟 ∫ 𝒖 𝒅𝒗 = 𝒖𝒗 − ∫ 𝒗 𝒅𝒖
This is known as the integration by parts formula and provides a method of integrating such
products of simple functions as ∫ 𝑥𝑒 𝑥 𝑑𝑥, ∫ 𝑡 sin 𝑡 𝑑𝑡, ∫ 𝑒 𝜃 cos 𝜃 𝑑𝜃 𝑎𝑛𝑑 ∫ 𝑥 ln 𝑥 𝑑𝑥.
Given a product of two terms to integrate the initial choice is: ‘which part to make equal to u’
and ‘which part to make equal to v’. The choice must be such that the ‘u part’ becomes a
constant after successive differentiation and the ‘dv part’ can be integrated from standard
integrals. Invariable, the following rule holds: If a product to be integrated contains an algebraic
term (such as x, 𝑡 2 or 3θ) then this term is chosen as the u part. An exception
to this rule is when a ‘lnx’ term is involved; in this case lnx is chosen as the ‘upart’.
As a general guide, the acronym LIATE is used being abbreviation for the following functions:
L – Logarithmic function
I – Inverse functions
A – Algebraic functions
T – Trigonometric functions
E – Exponential Functions
The preceding function is taken as u while the remaining terms and dx form dv.
Examples
1. Determine ∫ 𝑥 cos 𝑥 𝑑𝑥.
46
From the integration by parts formula,
∫ 𝑢 𝑑𝑣 = 𝑢𝑣 − ∫ 𝑣 𝑑𝑢
𝑑𝑢
Let u = x, from which 𝑑𝑥 = 1 𝑖. 𝑒. 𝑑𝑢 = 𝑑𝑥. 𝐿𝑒𝑡 𝑑𝑣 = cos 𝑥𝑑𝑥, 𝑓𝑟𝑜𝑚 𝑤ℎ𝑖𝑐ℎ 𝑣 = ∫ cos 𝑥 𝑑𝑥
= sin 𝑥.
Expressions for u, du and v are now substituted into the ‘by parts’ formula as shown below.
∫ 𝑢 𝑑𝑣
=
𝑢𝑣
−
∫ 𝑣 𝑑𝑢
∫ 𝑥 cos 𝑥 𝑑𝑥 = (𝑥) (sin 𝑥) − ∫(sin 𝑥)( 𝑑𝑥)
i.e.
∫ 𝑥 cos 𝑥 𝑑𝑥 = 𝑥 sin 𝑥 − (− cos 𝑥) + 𝑐
= 𝒙 𝐬𝐢𝐧 𝒙 + 𝐜𝐨𝐬 𝒙 + 𝒄
[This result may be checked by differentiating the right hand side]
2. Find∫ 3𝑡𝑒 2𝑡 𝑑𝑡.
Let u = 3t, from which
1
2
𝑑𝑢
𝑑𝑡
= 3 𝑖. 𝑒. 𝑑𝑢 = 3𝑑𝑡. 𝐿𝑒𝑡 𝑑𝑣 = 𝑒 2𝑡 𝑑𝑡, 𝑓𝑟𝑜𝑚 𝑤ℎ𝑖𝑐ℎ 𝑣 = ∫ 𝑒 2𝑡 𝑑𝑡 =
𝑒 2𝑡 .
Substituting into
∫ 𝑢 𝑑𝑣 = 𝑢𝑣 − ∫ 𝑣 𝑑𝑢
gives:
1
1
3
3
3
3 1
∫ 3𝑡𝑒 2𝑡 𝑑𝑡 = (3𝑡) ( 𝑒 2𝑡 ) − ∫ ( 𝑒 2𝑡 ) (3𝑑𝑡) = 𝑡𝑒 2𝑡 − ∫ 𝑒 2𝑡 𝑑𝑡 = 𝑡𝑒 2𝑡 − ( 𝑒 2𝑡 ) + 𝑐
2
2
2
2
2
2 2
=
𝟑
3 2𝑡 3 𝑒 2𝑡
𝑡𝑒 − ( ) + 𝑐
2
2 2
𝟏
Hence∫ 𝟑𝒕𝒆𝟐𝒕 𝒅𝒕 = 𝟐 𝒆𝟐𝒕 (𝒕 − 𝟐) + 𝒄, which may be checked by differentiating.
47
𝜋
3.Evaluate ∫02 2𝜃 sin 𝜃 𝑑𝜃.
𝐿𝑒𝑡 𝑢 = 2𝜃, 𝑓𝑟𝑜𝑚 𝑤ℎ𝑖𝑐ℎ,
𝑑𝑢
= 2, 𝑖. 𝑒. 𝑑𝑢 = 2𝑑𝜃 𝑎𝑛𝑑 𝑑𝑣 = sin 𝜃𝑑𝜃, 𝑓𝑟𝑜𝑚 𝑤ℎ𝑖𝑐ℎ
𝑑𝜃
𝑣 = ∫ sin 𝜃 𝑑𝜃 = − cos 𝜃
Substituting into ∫ 𝑢 𝑑𝑣 = 𝑢𝑣 − ∫ 𝑣 𝑑𝑢 gives:
∫ 2𝜃 sin 𝜃 𝑑𝜃 = (2𝜃)(− cos 𝜃) − ∫(− cos 𝜃)(2𝑑𝜃) = −2𝜃 cos 𝜃 + 2 ∫ cos 𝜃 𝑑𝜃
= −2𝜃 cos 𝜃 + 2 sin 𝜃 + 𝑐
𝜋
𝜋
𝜋
𝜋
𝜋
Hence ∫02 2𝜃 sin 𝜃 𝑑𝜃 = [−2𝜃 cos 𝜃 + 2 sin 𝜃] 2 = [−2 ( 2 ) cos 2 + 2 sin 2 ] − [0 + 2 sin 0]
0
= (−0 + 2) − (0 + 0) = 2 𝑠𝑖𝑛𝑐𝑒 cos
𝜋
𝜋
= 0 𝑎𝑛𝑑 sin = 1.
2
2
1
4.Evaluate ∫0 5𝑥𝑒 4𝑥 𝑑𝑥, correct to 3 significant figures.
𝑑𝑢
Let u = 5x, from which 𝑑𝑥 = 5, i.e. du = 5dx and let 𝑑𝑣 = 𝑒 4𝑥 𝑑𝑥, from which, 𝑣 = ∫ 𝑒 4𝑥 𝑑𝑥
1
= 𝑒 4𝑥
4
Substituting into ∫ 𝑢 𝑑𝑣 = 𝑢𝑣 − ∫ 𝑣 𝑑𝑢 gives:
1 4𝑥
𝑒 4𝑥
5 4𝑥 5
5 4𝑥 5 𝑒 4𝑥
4𝑥
∫ 5𝑥𝑒 𝑑𝑥 = (5𝑥) ( 𝑒 ) − ∫ ( ) (5𝑑𝑥) = 𝑥𝑒 − ∫ 𝑒 𝑑𝑥 = 𝑥𝑒 − ( ) + 𝑐
4
4
4
4
4
4 4
5
1
= 𝑒 4𝑥 (𝑥 − ) + 𝑐
4
4
4𝑥
1
5
1
Hence∫0 5𝑥𝑒 4𝑥 𝑑𝑥 = [4 𝑒 4𝑥 (𝑥 − 4)]
5
1
5
1
15
15
1
= [4 𝑒 4 (1 − 4)] − [4 𝑒 0 (0 − 4)] = (16 𝑒 4 ) — 16
0
= 51.186 − 0.313 = 51.499 = 𝟓𝟏. 𝟓, correct to 3 significant figures
5.Determine ∫ 𝑥 2 sin 𝑥 𝑑𝑥.
𝑑𝑢
Let u = 𝑥 2 , from which, 𝑑𝑥 = 2𝑥, i.e. du = 2xdx, and let dv = sinx dx, from which,
𝑣 = ∫ sin 𝑥 𝑑𝑥 = − cos 𝑥
48
Substituting into ∫ 𝑢 𝑑𝑣 = 𝑢𝑣 − ∫ 𝑣 𝑑𝑢 gives:
∫ 𝑥 2 sin 𝑥 𝑑𝑥 = (𝑥 2 )(− cos 𝑥) − ∫(− cos 𝑥)(2𝑥 𝑑𝑥) = −𝑥 2 cos 𝑥 + 2 ∫ xcos 𝑥 𝑑𝑥
The integral, ∫ xcos 𝑥 𝑑𝑥, is not a ‘standard integral’ and it can only be determined by using the
integration by parts formula again.
From example 1, ∫ xcos 𝑥 𝑑𝑥 = 𝑥 sin 𝑥 + cos 𝑥
Hence ∫ 𝑥 2 sin 𝑥 𝑑𝑥 = −𝑥 2 cos 𝑥 + 2(𝑥 sin 𝑥 + cos 𝑥) + 𝑐
= −𝑥 2 cos 𝑥 + 2𝑥 sin 𝑥 + 2 cos 𝑥 + 𝑐 = (𝟐 − 𝒙𝟐 ) 𝐜𝐨𝐬 𝒙 + 𝟐𝒙 𝐬𝐢𝐧 𝒙 + 𝒄
In general, if the algebraic term of a product is of power n, then the integration by parts formula
is applied n times.
6.Find ∫ 𝑥 ln 𝑥 𝑑𝑥.
𝑑𝑢
1
The logarithmic function is chosen as the ‘u part’. Thus when u = lnx, then 𝑑𝑥 = 𝑥 , i.e. 𝑑𝑢 =
Letting 𝑑𝑣 = 𝑥𝑑𝑥 gives 𝑣 = ∫ 𝑥 𝑑𝑥 =
𝑥2
2
Substituting into ∫ 𝑢 𝑑𝑣 = 𝑢𝑣 − ∫ 𝑣 𝑑𝑢 gives:
𝑥2
𝑥 2 𝑑𝑥 𝑥 2
1
𝑥2
1 𝑥2
∫ 𝑥 ln 𝑥 𝑑𝑥 = (ln 𝑥) ( ) − ∫ ( )
= ln 𝑥 − ∫ 𝑥 𝑑𝑥 = ln 𝑥 − ( ) + 𝑐
2
2 𝑥
2
2
2
2 2
Hence ∫ 𝑥 ln 𝑥 𝑑𝑥 =
𝒙𝟐
𝟐
𝟏
(𝐥𝐧 𝒙 − 𝟐) + 𝒄
Practice
1.Determine ∫ ln 𝑥 𝑑𝑥
2.Find ∫ 𝑥 3 ln 𝑥 𝑑𝑥
7. Determine ∫ 𝑥 2 𝑒 𝑥 𝑑𝑥.
𝑑𝑢
Let 𝑢 = 𝑥 2 , 𝑑𝑥 = 2𝑥, 𝑑𝑢 = 2𝑥𝑑𝑥 𝑎𝑛𝑑 𝑑𝑣 = 𝑒 𝑥 𝑑𝑥 𝑜𝑟 𝑣 = ∫ 𝑒 𝑥 𝑑𝑥 = 𝑒 𝑥 .
Substituting into ∫ 𝑢 𝑑𝑣 = 𝑢𝑣 − ∫ 𝑣 𝑑𝑢 gives:
∫ 𝑥 2 𝑒 𝑥 𝑑𝑥 = 𝑥 2 𝑒 𝑥 − ∫ 𝑒 𝑥 (2𝑥𝑑𝑥) = 𝑥 2 𝑒 𝑥 − 2 ∫ 𝑥𝑒 𝑥 𝑑𝑥
49
𝑑𝑥
𝑥
The integral, ∫ 𝑥𝑒 𝑥 𝑑𝑥, is not a ‘standard integral’ and it can only be determined by using the
integration by parts formula again.
𝑑𝑢
Let 𝑢 = 𝑥, 𝑑𝑥 = 1, 𝑑𝑢 = 𝑑𝑥 𝑎𝑛𝑑 𝑑𝑣 = 𝑒 𝑥 𝑑𝑥 𝑜𝑟 𝑣 = ∫ 𝑒 𝑥 𝑑𝑥 = 𝑒 𝑥 .
∫ 𝑥𝑒 𝑥 𝑑𝑥 = 𝑥𝑒 𝑥 − ∫ 𝑒 𝑥 𝑑𝑥 = 𝑥𝑒 𝑥 − 𝑒 𝑥
Therefore∫ 𝑥 2 𝑒 𝑥 𝑑𝑥 = 𝑥 2 𝑒 𝑥 − 2(𝑥𝑒 𝑥 − 𝑒 𝑥 ) + 𝑐 = (𝒙𝟐 − 𝟐𝒙 − 𝟏)𝒆𝒙 + 𝒄
Note that the integration by parts formula is applied twice
Reduction Formulae
When using integration by parts, an
integral such as∫ 𝑥 2 𝑒 𝑥 𝑑𝑥 requires integration by parts twice. Similarly, ∫ 𝑥 3 𝑒 𝑥 𝑑𝑥 requires
integration by parts three times. Thus, integrals like ∫ 𝑥 5 𝑒 𝑥 𝑑𝑥, ∫ 𝑥 6 cos 𝑥 𝑑𝑥 𝑎𝑛𝑑 ∫ 𝑥 8 sin 𝑥 𝑑𝑥
for example, would take a long time to determine using integration by parts. Reduction formulae
provide a quicker method for determining such integrals and the method is demonstrated below.
Using reduction formulae for integrals of the form∫ 𝒙𝒏 𝒆𝒙 𝒅𝒙
To determine∫ 𝑥 𝑛 𝑒 𝑥 𝑑𝑥 using integration by parts, let 𝑢 = 𝑥 𝑛 from which,
𝑑𝑢
= 𝑛𝑥 𝑛−1 𝑎𝑛𝑑 𝑑𝑢 = 𝑛𝑥 𝑛−1 𝑑𝑥; 𝑑𝑣 = 𝑒 𝑥 𝑑𝑥 𝑓𝑟𝑜𝑚 𝑤ℎ𝑖𝑐ℎ 𝑣 = ∫ 𝑒 𝑥 𝑑𝑥 = 𝑒 𝑥
𝑑𝑥
Thus, ∫ 𝑥 𝑛 𝑒 𝑥 𝑑𝑥 = 𝑥 𝑛 𝑒 𝑥 − ∫ 𝑒 𝑥 𝑛𝑥 𝑛−1 𝑑𝑥, using the integration by parts formula, i.e.
∫ 𝑥 𝑛 𝑒 𝑥 𝑑𝑥 = 𝑥 𝑛 𝑒 𝑥 − 𝑛 ∫ 𝑥 𝑛−1 𝑒 𝑥 𝑑𝑥.
The integral on the far right is seen to be of the same form as the integral on the left-hand side,
except that n has been replaced by n−1.
Thus, if we let, ∫ 𝑥 𝑛 𝑒 𝑥 𝑑𝑥 = 𝐼𝑛 , 𝑡ℎ𝑒𝑛 ∫ 𝑥 𝑛−1 𝑒 𝑥 𝑑𝑥 = 𝐼𝑛−1 .
Hence ∫ 𝑥 𝑛 𝑒 𝑥 𝑑𝑥 = 𝑥 𝑛 𝑒 𝑥 − 𝑛 ∫ 𝑥 𝑛−1 𝑒 𝑥 𝑑𝑥 can be written as :
𝐼𝑛 = 𝑥 𝑛 𝑒 𝑥 − 𝑛𝐼𝑛−1 … (1)
Equation (1) is an example of a reduction formula since it expresses an integral in n in terms of
the same integral in n−1.
Examples
50
1. Determine∫ 𝑥 2 𝑒 𝑥 𝑑𝑥 using a reduction formula.
Using equation (1) with n = 2 gives:
∫ 𝑥 2 𝑒 𝑥 𝑑𝑥 = 𝐼2 = 𝑥 2 𝑒 𝑥 − 2𝐼1 ;
and 𝐼1 = 𝑥1 𝑒 𝑥 − 1𝐼0
𝐼0 = ∫ 𝑥 0 𝑒 𝑥 𝑑𝑥 = ∫ 𝑒 𝑥 𝑑𝑥 = 𝑒 𝑥 + 𝑐 ′
Hence 𝐼2 = 𝑥 2 𝑒 𝑥 − 2[𝑥1 𝑒 𝑥 − 1𝐼0 ]
= 𝑥 2 𝑒 𝑥 − 2[𝑥1 𝑒 𝑥 − 1(𝑒 𝑥 + 𝑐 ′ )] = 𝑥 2 𝑒 𝑥 − 2𝑥𝑒 𝑥 + 2𝑒 𝑥 + 𝑐
Thus∫ 𝑥 2 𝑒 𝑥 𝑑𝑥 = (𝒙𝟐 − 𝟐𝒙 − 𝟏)𝒆𝒙 + 𝒄
Notice that the constant of integration is added at the last step with indefinite integrals.
2. Use a reduction formula to determine ∫ 𝑥 3 𝑒 𝑥 𝑑𝑥.
From equation (1), 𝐼𝑛 = 𝑥 𝑛 𝑒 𝑥 − 𝑛𝐼𝑛−1 .
Hence ∫ 𝑥 3 𝑒 𝑥 𝑑𝑥 = 𝐼3 = 𝑥 3 𝑒 𝑥 − 3𝐼2
𝐼2 = 𝑥 2 𝑒 𝑥 − 3𝐼1
𝐼1 = 𝑥1 𝑒 𝑥 − 1𝐼0
and 𝐼0 = ∫ 𝑥 0 𝑒 𝑥 𝑑𝑥 = ∫ 𝑒 𝑥 𝑑𝑥 = 𝑒 𝑥
Thus ∫ 𝑥 3 𝑒 𝑥 𝑑𝑥 = 𝑥 3 𝑒 𝑥 − 3[𝑥 2 𝑒 𝑥 − 2𝐼1 ] = 𝑥 3 𝑒 𝑥 − 3[𝑥 2 𝑒 𝑥 − 2(𝑥1 𝑒 𝑥 − 1𝐼0 )]
= 𝑥 3 𝑒 𝑥 − 3[𝑥 2 𝑒 𝑥 − 2(𝑥1 𝑒 𝑥 − 1𝑒 𝑥 )] = 𝑥 3 𝑒 𝑥 − 3𝑥 2 𝑒 𝑥 + 6(𝑥1 𝑒 𝑥 − 1𝑒 𝑥 )
= 𝑥 3 𝑒 𝑥 − 3𝑥 2 𝑒 𝑥 + 6(𝑥1 𝑒 𝑥 − 1𝑒 𝑥 ) = 𝑥 3 𝑒 𝑥 − 3𝑥 2 𝑒 𝑥 + 6𝑥𝑒 𝑥 − 6𝑒 𝑥
i.e. ∫ 𝑥 3 𝑒 𝑥 𝑑𝑥 = 𝑥 3 𝑒 𝑥 − 3𝑥 2 𝑒 𝑥 + 6𝑥𝑒 𝑥 − 6𝑒 𝑥 + 𝑐 = 𝒆𝒙 (𝒙𝟑 − 𝟑𝒙𝟐 + 𝟔𝒙 − 𝟔) + 𝒄
Using reduction formulae for integrals of the form∫ 𝒙𝒏 𝒄𝒐𝒔 𝒙 𝒅𝒙 𝒂𝒏𝒅 ∫ 𝒙𝒏 𝒔𝒊𝒏 𝒙 𝒅𝒙
(a) ∫ 𝒙𝒏 𝒄𝒐𝒔 𝒙 𝒅𝒙
Let 𝐼𝑛 = ∫ 𝑥 𝑛 cos 𝑥 𝑑𝑥 then, using integration by parts:
if 𝑢 = 𝑥 𝑛 𝑡ℎ𝑒𝑛
𝑑𝑢
𝑑𝑥
= 𝑛𝑥 𝑛−1 𝑎𝑛𝑑 𝑑𝑢 = 𝑛𝑥 𝑛−1 𝑑𝑥 𝑎𝑛𝑑 𝑖𝑓 𝑑𝑣 = cos 𝑥𝑑𝑥 𝑡ℎ𝑒𝑛 𝑣 = ∫ cos 𝑥 𝑑𝑥
= sin 𝑥
51
Using the integration by parts formula ∫ 𝑢 𝑑𝑣 = 𝑢𝑣 − ∫ 𝑣 𝑑𝑢,
𝐼𝑛 = 𝑥 𝑛 sin 𝑥 − ∫(sin 𝑥)𝑛𝑥 𝑛−1 𝑑𝑥 = 𝑥 𝑛 sin 𝑥 − 𝑛 ∫ 𝑥 𝑛−1 sin 𝑥 𝑑𝑥
Using integration by parts again, this time with 𝑢 = 𝑥 𝑛−1
𝑑𝑢
= (𝑛 − 1)𝑥 𝑛−2 , 𝑎𝑛𝑑 𝑑𝑣 = sin 𝑥𝑑𝑥 𝑓𝑟𝑜𝑚 𝑤ℎ𝑖𝑐ℎ 𝑣 = ∫ sin 𝑥 𝑑𝑥 = − cos 𝑥
𝑑𝑥
Hence 𝐼𝑛 = 𝑥 𝑛 sin 𝑥 − 𝑛[𝑥 𝑛−1 (− cos 𝑥) − ∫(− cos 𝑥)(𝑛 − 1)𝑥 𝑛−2 𝑑𝑥]
= 𝑥 𝑛 sin 𝑥 + 𝑛𝑥 𝑛−1 cos 𝑥 − 𝑛(𝑛 − 1) ∫ 𝑥 𝑛−2 cos 𝑥 𝑑𝑥
i.e. 𝑰𝒏 = 𝒙𝒏 𝐬𝐢𝐧 𝒙 + 𝒏𝒙𝒏−𝟏 𝐜𝐨𝐬 𝒙 − 𝒏(𝒏 − 𝟏)𝑰𝒏−𝟐 … (2)
Example
Use a reduction formula to determine ∫ 𝑥 2 cos 𝑥 𝑑𝑥.
Using the reduction formula of equation (2):
∫ 𝑥 2 cos 𝑥 𝑑𝑥 = 𝐼2 = 𝑥 2 sin 𝑥 + 2𝑥1 cos 𝑥 − 2(1)𝐼0 𝑎𝑛𝑑 𝐼0 = ∫ 𝑥 0 cos 𝑥 𝑑𝑥 = ∫ cos 𝑥 𝑑𝑥
= 𝑠𝑖𝑛𝑥
Hence ∫ 𝑥 2 cos 𝑥 𝑑𝑥 = 𝒙𝟐 𝐬𝐢𝐧 𝒙 + 𝟐𝒙 𝐜𝐨𝐬 𝒙 − 𝟐 𝐬𝐢𝐧 𝒙 + 𝒄
(b) ∫ 𝒙𝒏 𝒔𝒊𝒏 𝒙 𝒅𝒙
Let 𝐼𝑛 = ∫ 𝑥 𝑛 sin 𝑥 𝑑𝑥 then, using integration by parts:
if 𝑢 = 𝑥 𝑛 𝑡ℎ𝑒𝑛
𝑑𝑢
𝑑𝑥
= 𝑛𝑥 𝑛−1 𝑎𝑛𝑑 𝑑𝑢 = 𝑛𝑥 𝑛−1 𝑑𝑥 𝑎𝑛𝑑 𝑖𝑓 𝑑𝑣 = sin 𝑥𝑑𝑥 𝑡ℎ𝑒𝑛 𝑣 = ∫ sin 𝑥 𝑑𝑥
= − cos 𝑥
Using the integration by parts formula ∫ 𝑢 𝑑𝑣 = 𝑢𝑣 − ∫ 𝑣 𝑑𝑢,
∫ 𝑥 𝑛 sin 𝑥 𝑑𝑥 = 𝐼𝑛 = 𝑥 𝑛 (−cos 𝑥) − ∫(−cos 𝑥)𝑛𝑥 𝑛−1 𝑑𝑥 = −𝑥 𝑛 cos 𝑥 + 𝑛 ∫ 𝑥 𝑛−1 cos 𝑥 𝑑𝑥
Using integration by parts again, this time with 𝑢 = 𝑥 𝑛−1
𝑑𝑢
= (𝑛 − 1)𝑥 𝑛−2 , 𝑎𝑛𝑑 𝑑𝑣 = cos 𝑥𝑑𝑥 𝑓𝑟𝑜𝑚 𝑤ℎ𝑖𝑐ℎ 𝑣 = ∫ cos 𝑥 𝑑𝑥 = sin 𝑥
𝑑𝑥
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Hence 𝐼𝑛 = −𝑥 𝑛 cos 𝑥 + 𝑛[𝑥 𝑛−1 (sin 𝑥) − ∫(sin 𝑥)(𝑛 − 1)𝑥 𝑛−2 𝑑𝑥]
= −𝑥 𝑛 cos 𝑥 + 𝑛𝑥 𝑛−1 sin 𝑥 − 𝑛(𝑛 − 1) ∫ 𝑥 𝑛−2 sin 𝑥 𝑑𝑥
i.e. 𝑰𝒏 = −𝒙𝒏 𝐜𝐨𝐬 𝒙 + 𝒏𝒙𝒏−𝟏 𝐬𝐢𝐧 𝒙 − 𝒏(𝒏 − 𝟏)𝑰𝒏−𝟐 … (3)
Example
Use a reduction formula to determine ∫ 𝑥 3 sin 𝑥 𝑑𝑥.
Using equation (3),
∫ 𝑥 3 sin 𝑥 𝑑𝑥 = 𝐼3 = −𝑥 3 cos 𝑥 + 3𝑥 2 sin 𝑥 − (3)(2)𝐼1 𝑎𝑛𝑑 𝐼1
= −𝑥1 cos 𝑥 + 1𝑥 0 sin 𝑥 = −𝑥𝑐𝑜𝑠 𝑥 + sin 𝑥
Hence ∫ 𝑥 3 sin 𝑥 𝑑𝑥 = −𝑥 3 cos 𝑥 + 3𝑥 2 sin 𝑥 − 6[−𝑥𝑐𝑜𝑠 𝑥 + sin 𝑥]
= −𝒙𝟑 𝐜𝐨𝐬 𝒙 + 𝟑𝒙𝟐 𝐬𝐢𝐧 𝒙 + 𝟔𝒙 𝐜𝐨𝐬 𝒙 − 𝟔 𝐬𝐢𝐧 𝒙 + 𝒄
53