(1) The document discusses various integration techniques including: review of integral formulas, integration by parts, trigonometric integrals involving products of sines and cosines, trigonometric substitutions, and integration of rational functions using partial fractions.
(2) Examples are provided to demonstrate each technique, such as using integration by parts to evaluate integrals of the form ∫udv, using trigonometric identities to reduce powers of trigonometric functions, and using partial fractions to break down rational functions into simpler fractions.
(3) The key techniques discussed are integration by parts, trigonometric substitutions to transform integrals involving quadratic expressions into simpler forms, and partial fractions to decompose rational functions for integration. Various examples illustrate the
Integration by substitution is the chain rule in reverse.
NOTE: the final location is section specific. Section 1 (morning) is in SILV 703, Section 11 (afternoon) is in CANT 200
Integration by substitution is the chain rule in reverse.
NOTE: the final location is section specific. Section 1 (morning) is in SILV 703, Section 11 (afternoon) is in CANT 200
Integration Made Easy!
The derivative of a function can be geometrically interpreted as the slope of the curve of the mathematical function f(x) plotted as a function of x. But its implications for the modeling of nature go far deeper than this simple geometric application might imply. After all, you can see yourself drawing finite triangles to discover slope, so why is the derivative so important? Its importance lies in the fact that many physical entities such as velocity, acceleration, force and so on are defined as instantaneous rates of change of some other quantity. The derivative can give you a precise intantaneous value for that rate of change and lead to precise modeling of the desired quantity.
Basic concepts of integration, definite and indefinite integrals,properties of definite integral, problem based on properties,method of integration, substitution, partial fraction, rational , irrational function integration, integration by parts, reduction formula, improper integral, convergent and divergent of integration
this is the ppt on application of integrals, which includes-area between the two curves , volume by slicing , disk method , washer method, and volume by cylindrical shells,.
this is made by dhrumil patel and harshid panchal.
Integration Made Easy!
The derivative of a function can be geometrically interpreted as the slope of the curve of the mathematical function f(x) plotted as a function of x. But its implications for the modeling of nature go far deeper than this simple geometric application might imply. After all, you can see yourself drawing finite triangles to discover slope, so why is the derivative so important? Its importance lies in the fact that many physical entities such as velocity, acceleration, force and so on are defined as instantaneous rates of change of some other quantity. The derivative can give you a precise intantaneous value for that rate of change and lead to precise modeling of the desired quantity.
Basic concepts of integration, definite and indefinite integrals,properties of definite integral, problem based on properties,method of integration, substitution, partial fraction, rational , irrational function integration, integration by parts, reduction formula, improper integral, convergent and divergent of integration
this is the ppt on application of integrals, which includes-area between the two curves , volume by slicing , disk method , washer method, and volume by cylindrical shells,.
this is made by dhrumil patel and harshid panchal.
Integration is a part of Calculus.
This is just a short presentation on Integration.
It may help you out to complete your academic presentation.
Thank You
We all have good and bad thoughts from time to time and situation to situation. We are bombarded daily with spiraling thoughts(both negative and positive) creating all-consuming feel , making us difficult to manage with associated suffering. Good thoughts are like our Mob Signal (Positive thought) amidst noise(negative thought) in the atmosphere. Negative thoughts like noise outweigh positive thoughts. These thoughts often create unwanted confusion, trouble, stress and frustration in our mind as well as chaos in our physical world. Negative thoughts are also known as “distorted thinking”.
This is a presentation by Dada Robert in a Your Skill Boost masterclass organised by the Excellence Foundation for South Sudan (EFSS) on Saturday, the 25th and Sunday, the 26th of May 2024.
He discussed the concept of quality improvement, emphasizing its applicability to various aspects of life, including personal, project, and program improvements. He defined quality as doing the right thing at the right time in the right way to achieve the best possible results and discussed the concept of the "gap" between what we know and what we do, and how this gap represents the areas we need to improve. He explained the scientific approach to quality improvement, which involves systematic performance analysis, testing and learning, and implementing change ideas. He also highlighted the importance of client focus and a team approach to quality improvement.
Read| The latest issue of The Challenger is here! We are thrilled to announce that our school paper has qualified for the NATIONAL SCHOOLS PRESS CONFERENCE (NSPC) 2024. Thank you for your unwavering support and trust. Dive into the stories that made us stand out!
Palestine last event orientationfvgnh .pptxRaedMohamed3
An EFL lesson about the current events in Palestine. It is intended to be for intermediate students who wish to increase their listening skills through a short lesson in power point.
Ethnobotany and Ethnopharmacology:
Ethnobotany in herbal drug evaluation,
Impact of Ethnobotany in traditional medicine,
New development in herbals,
Bio-prospecting tools for drug discovery,
Role of Ethnopharmacology in drug evaluation,
Reverse Pharmacology.
Synthetic Fiber Construction in lab .pptxPavel ( NSTU)
Synthetic fiber production is a fascinating and complex field that blends chemistry, engineering, and environmental science. By understanding these aspects, students can gain a comprehensive view of synthetic fiber production, its impact on society and the environment, and the potential for future innovations. Synthetic fibers play a crucial role in modern society, impacting various aspects of daily life, industry, and the environment. ynthetic fibers are integral to modern life, offering a range of benefits from cost-effectiveness and versatility to innovative applications and performance characteristics. While they pose environmental challenges, ongoing research and development aim to create more sustainable and eco-friendly alternatives. Understanding the importance of synthetic fibers helps in appreciating their role in the economy, industry, and daily life, while also emphasizing the need for sustainable practices and innovation.
How to Split Bills in the Odoo 17 POS ModuleCeline George
Bills have a main role in point of sale procedure. It will help to track sales, handling payments and giving receipts to customers. Bill splitting also has an important role in POS. For example, If some friends come together for dinner and if they want to divide the bill then it is possible by POS bill splitting. This slide will show how to split bills in odoo 17 POS.
1. INTEGRATION TECHNIQUES
1. Review of formulas and techniques
Summary of known integral formulas.
ˆ
xr+1
x dx =
+ c, for r = −1 (power rule)
r+1
ˆ
r
ˆ
ˆ
sin x dx = − cos x + c
cos x dx = sin x + c
ˆ
ˆ
2
sec x dx = tan x + c
sec x tan x dx = sec x + c
ˆ
ˆ
2
csc x dx = − cot x + c
csc x cot x dx = − csc x + c
ˆ
ˆ
x
e−x dx = −e−x + c
x
e dx = e + c
ˆ
ˆ
√
tan x dx = − ln | cos x| + c
ˆ
1
dx = ln |x| + c for x = 0
x
ˆ
1
dx = tan−1 x + c
2
1+x
1
dx = sin−1 x + c
2
1−x
1
√
dx = sec−1 x + c
2−1
|x| x
Example 1.1. (A Simple Substitution)
ˆ
Evaluate sin(ax)dx, for a = 0.
Example 1.2. (Generalizing a Basic Integration Rule)
ˆ
1
Evaluate
dx, for a = 0.
2 + x2
a
Example 1.3. (An Integrand That Must Be Expanded)
ˆ
Evaluate (x2 − 5)2 dx.
Example 1.4. (An Integral Where We Must Complete the Square)
ˆ
1
√
dx.
Evaluate
−5 + 6x − x2
Example 1.5. (An Integral Requiring Some Imagination)
ˆ
4x + 1
Evaluate
dx.
2 + 4x + 10
2x
1
2. 2
INTEGRATION TECHNIQUES
2. Integration by Parts
Remember that if u and v are differentiable functions of x, then u · v is also differentiable and
d
d
d
(u · v) = u · v + v · u.
dx
dx
dx
Using differential forms this may be written as
d(uv) = udv + vdu.
Taking the integral of the both sides of (1), we obtain
ˆ
ˆ
ˆ
d(uv) = udv + vdu.
Note that
´
(1)
(2)
d(uv) = uv. Therefore (2) can be written as
ˆ
ˆ
uv = udv + vdu,
or, equivalently,
ˆ
ˆ
udv = uv −
vdu.
Remark. This technique requires the separation of the integral into two parts as u and dv
(which includes the differential of integral, say dx) such that
• u gets simpler when differentiated.
• dv is easy to integrate.
Example 2.1. Evaluate the following integrals.
ˆ
(1)
x sin x dx
ˆ
(2)
x2 ex dx
ˆ
(3)
ex cos x dx
ˆ
(4)
x ln x dx
ˆ
(5)
ln x dx
ˆ
(6)
sin−1 x dx
ˆ
2
(7)
x3 ex dx
ˆ
(8)
x sec2 x dx
ˆ
(9)
x3x dx
ˆ
(10)
cos(ln x) dx
ˆ
(11)
x4 ex dx
3. INTEGRATION TECHNIQUES
ˆ
(12)
(13)
(14)
(15)
(16)
(17)
(18)
(19)
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
3
√
ln x dx
√
e
x
dx
x2 sin x dx
xe2x dx
xe−x dx
e2x cos 3x dx
tan−1 x dx
2
ˆ
x5 e−x dx
−1
esin x dx
ˆ
√
(21)
x 1 − x dx
ˆ 2
(22)
x3 ln x dx
(20)
1
3. Trigonometric Integrals
3.1. Products of Powers of Sines and Cosines. If we are given an integral of the form
ˆ
sinm x cosn x dx, where m, n ≥ 0,
There are three possible cases:
Case 1. If m is odd, we write m as 2k + 1 and use the identity sin2 x = 1 − cos2 x to obtain
sinm x = sin2k+1 x = (sin2 x)k sin x = (1 − cos2 x)k sin x,
and, then
ˆ
ˆ
m
n
sin x cos x dx =
(1 − cos2 x)k cosn x sin x dx.
Using the substitution u = cos x, we get
ˆ
ˆ
ˆ
m
n
2
k
n
sin x cos x dx = (1 − cos x) cos x sin x dx = − (1 − u2 )k un du.
Now, the integral is easy to evaluate.
Case 2. If m is even and n is odd, we write n as 2k + 1 and use the identity cos2 x = 1 − sin2 x
and repeat what we have done in Case 1 changing the places of sine and cosine.
Case 3. If both m and n are even, we substitute
1 − cos 2x
1 + cos 2x
, cos2 x =
2
2
to reduce the integrand to one in lower powers of cos 2x.
sin2 x =
4. 4
INTEGRATION TECHNIQUES
3.2. Integrals of Powers of tan x and sec x. We know how to integrate the tangent and
secant and their squares. To integrate higher powers we use the identities tan2 x = sec2 x − 1
and sec2 x = tan2 x + 1, and integrate by parts when necessary to reduce the higher powers to
lower powers.
3.3. Products of Sines and Cosines. To evaluate the integrals of the form
ˆ
ˆ
ˆ
sin mx sin nx dx,
sin mx cos nx dx, and
cos mx cos nx dx,
we may apply the method of integration by parts, but two such integrations are required in
each case. So, it is better to use the identities
1
sin mx sin nx = (cos(m − n)x − cos(m + n)x)
2
1
sin mx cos nx = (sin(m − n)x + cos(m + n)x)
2
1
cos mx cos nx = (cos(m − n)x + cos(m + n)x)
2
Example 3.1. Evaluate the following integrals.
ˆ
(1)
cos4 x sin x dx
ˆ
(2)
cos4 x sin3 x dx
ˆ √
(3)
sin x cos5 x dx
ˆ
(4)
cos3 x dx
ˆ
(5)
sin3 x cos7 x dx
ˆ
(6)
sin2 x dx
ˆ
(7)
cos4 x dx
ˆ
(8)
sin2 x cos2 x dx
ˆ
(9)
tan3 x sec3 x dx
ˆ
(10)
tan2 x dx
ˆ
(11)
tan2 x sec4 x dx
ˆ
(12)
tan3 x dx
ˆ
(13)
tan3 x sec5 x dx
5. INTEGRATION TECHNIQUES
5
ˆ
(14)
sec x dx
ˆ √
tan x sec4 x dx
(15)
ˆ
(16)
cot4 x dx
ˆ
(17)
csc x dx
ˆ
(18)
cot3 x csc3 x dx
ˆ
sin5 x
√
(19)
dx
cos x
3.4. Trigonometric Substitutions. Trigonometric substitutions can be effective in transforming complicated integrals involving a2 − x2 , a2 + x2 and x2 − a2 into integrals we can
evaluate directly.
The following table gives the suitable trigonometric substitution in each case and the trigonometric identity to be used.
Expression Trigonometric Substitution Related Trigonometric Identity
a2 + x 2
x = a tan θ
1 + tan2 θ = sec2 θ
2
2
a −x
x = a sin θ
1 − sin2 θ = cos2 θ
x 2 − a2
x = a sec θ
sec2 θ − 1 = tan2 θ
Remark. We want any substitution we use in an integration to be reversible so that we can
change back to the original variable afterward. For example, if x = a tan θ, we want to be able
x
after the integration takes place. As we know, the functions in these
to set θ = tan−1
a
substitution have inverses only for selected values of θ. For reversibility,
x
π
π
x = a tan θ requires θ = tan−1
with − < θ < ,
a
2
2
π
π
≤θ≤ ,
2
2
0 ≤ θ < π if x ≥ 1,
2
a
−1 x
x = a sec θ requires θ = sec
with
a
π < θ ≤ π if x ≤ −1,
2
a
To simplify calculations with the substitution x = a sec θ, we will restrict its use to integrals in
√
x
π
which ≥ 1. This will place θ in 0,
and make tan θ ≥ 0. We will then have x2 − a2 =
a
2
√
a2 tan2 θ = |a tan θ| = a tan θ, free of absolute values, provided that a > 0.
x = a sin θ
requires θ = sin−1
x
a
with −
Example 3.2.
(1) Evaluate the following integrals.
ˆ
dx
√
(a)
2
2
ˆ x 4−x
dx
√
(b)
2
ˆ √9 2+ x
x − 25
(c)
dx, for x ≥ 5.
x
6. 6
INTEGRATION TECHNIQUES
ˆ
(d)
ˆ
(e)
ˆ
(f)
ˆ
(g)
ˆ
(h)
(i)
(j)
ˆ
ˆ
ˆ
dx
(1 + x2 )2
√
4 − x2 dx
√
x2 − 1
dx
x
dx
√
2 x2 − 2
x
√
9x2 − 4
dx
x
dx
2 + 2x + 10
x
√
ex 1 − e2x dx
√
cos x
dx
1 + sin2 x
dx
(l)
2 + 4x + 5)2
ˆ (4x
2x + 2
(m)
dx
2 − 4x + 8
ˆ x
5x + 3
dx
(n)
2 − 4x + 8
ˆ x
x2
(o)
5 dx
(x2 − 1) 2
(2) Find the volume of the solid generated by revolving about the x-axis the region in the
2
first quadrant enclosed by the coordinate axes, the curve y =
, and the line x = 1.
1 + x2
(k)
ˆ
4. Integration of Rational Functions Using Partial Fractions
The method based on the fact that “every rational function can be written as a sum of simpler
fractions that we can integrate with the techniques we already know”.
General Description of the Method of Partial Fractions. To write a rational function
P (x)
as a sum of partial fractions, do the following:
Q(x)
(1) The degree of P (x) must be less than the degree of Q(x). (Otherwise, divide P (x) by
Q(x) and work on the remainder.)
(2) Factor out the polynomial Q(x).
• For each linear factor, assign the sum of m partial fractions to this factor where m
is the exponent of that linear factor as follows:
A1
A2
Am
+
+ ··· +
2
x − r (x − r)
(x − r)m
• For each irreducible quadratic factor (a polynomial of second degree that cannot
be written as a product of linear factors), assign the sum of n partial fractions to
7. INTEGRATION TECHNIQUES
this factor where n is the exponent of that quadratic factor as:
B1 x + C1
B2 x + C2
Bn x + Cn
+ 2
+ ··· + 2
2 + px + q
2
x
(x + px + q)
(x + px + q)n
Several Examples. We assume that the degree of P (x) is less than the degree of Q(x).
(1) Distinct Linear Factors:
P (x)
A
B
C
P (x)
=
=
+
+
.
Q(x)
(x − r1 )(x − r2 )(x − r3 )
x − r1 x − r2 x − r3
(2) Repeated Linear Factors:
P (x)
P (x)
B
A
C
D
=
+
=
+
+
.
4
2
3
Q(x)
(x − r)
x − r (x − r)
(x − r)
(x − r)4
(3) Some distinct, some repeated linear factors:
P (x)
A B
C
D
E
F
P (x)
+
+
= 2
= + 2+
+
.
3
2
Q(x)
x (x − r1 )(x − r2 )
x x
x − r1 x − r2 (x − r2 )
(x − r3 )3
(4) Distinct irreducible quadratic factors:
P (x)
P (x)
Ax + B
Cx + D
= 2
= 2
+ 2
.
2+p x+q )
Q(x)
(x + p1 x + q1 )(x
x + p1 x + q 1 x + p2 x + q2
2
2
(5) Repeated irreducible quadratic factors:
P (x)
P (x)
Ax + B
Cx + D
Ex + F
= 2
= 2
+ 2
+ 2
.
3
2
Q(x)
(x + p1 x + q1 )
x + p1 x + q1 (x + p1 x + q1 )
(x + p1 x + q1 )3
(6) General Case:
P (x)
P (x)
=
2 (x2 + p x + q )2 (x2 + p x + q )
Q(x)
x(x − r)
1
1
2
2
A
B
C
Dx + E
Fx + G
Hx + I
= +
+
+ 2
+ 2
+ 2
.
2
2
x x − r (x − r)
x + p1 x + q1 (x + p1 x + q1 )
x + p2 x + q 2
Example 4.1. Evaluate the following integrals.
ˆ
1
(1)
dx
2+x−2
ˆ x 2
3x − 7x − 2
(2)
dx
x3 − x
ˆ
2x3 − 4x2 − 15x + 5
(3)
dx
x2 − 2x − 8
ˆ
5x2 + 20x + 6
(4)
dx
3
2
ˆ x 2 + 2x + x
2x − 5x + 2
(5)
dx
x3 + x
ˆ
5x2 + 6x + 2
dx
(6)
2
ˆ (x + 2)(x + 2x + 5)
3x − 1
(7)
dx
(x + 2)(x − 3)
7
9. INTEGRATION TECHNIQUES
9
Brief Summary of Integration Techniques.
ˆ
ˆ
• Integration by Substitution:
f (u(x))u (x) dx = f (u) du
What to look for
– Compositions of the form f (u(x)), where the integrand also contains u (x); for
example,
ˆ
ˆ
ˆ
2
2
2x cos(x ) dx = cos(x )2x dx = cos udu.
– Compositions of the form f (ax + b); for example,
ˆ
ˆ
x
u−1
√
√ du.
dx =
u
x+1
´
´
• Integration by Parts: u dv = uv − v du
What to look for: products of different types of functions: xn , cos x, ex ; for example,
ˆ
ˆ
2x cos x dx = x sin x − sin x dx.
• Trigonometric Substitutions:
What to look √
for:
– Terms like a2 − x2 : Let x = a sin θ, − π ≤ θ ≤ π , so that dx = a cos θ dθ and
2
2
√
a2 − x2 = a2 − a2 sin2 θ = a cos θ; for example,
ˆ
ˆ
x2
√
dx = sin2 θ dθ.
2
1−x
√
– Terms like x2 + a2 : Let x = a tan θ, − π < θ < π , so that dx = a sec2 θ dθ and
2
2
√
√
x2 + a2 = a2 tan2 θ + a2 = a sec θ; for example,
ˆ
ˆ
x3
√
dx = 27 tan3 θ sec θ dθ.
2+9
x
√
2 − a2 : Let x = a sec θ, for θ ∈ [0, π ) ∪ ( π , π], so that dx =
– Terms like x
2
2
√
√
a sec θ tan θ dθ and x2 − a2 = a2 sec2 θ − a2 = a tan θ; for example,
ˆ
ˆ
√
3
2 − 4 dx = 32
x x
sec4 θ tan2 θ dθ.
• Partial Fractions:
What to look for: rational functions, for example,
ˆ
ˆ
ˆ
x+2
x+2
A
B
dx =
dx =
+
2 − 4x + 3
x
(x − 1)(x − 3)
x−1 x−3
dx.
10. 10
INTEGRATION TECHNIQUES
5. Improper Integrals
ˆ
2
Example 5.1. Evaluate
−1
1
dx.
x2
When studying the definite integrals, we required two things. First, the domain of integration
(from a to b) [a, b], be finite. Second, function is finite on the domain of integration.
Question: What happens if the domain of integration is infinite? What if function becomes
infinite in the domain of integration?
Answer: Improper integration!
The integrals of type described above are called improper integrals. If the limits exist, we
evaluate them with the following definitions
(1) If f is continuous on [a, ∞), then
ˆ ∞
ˆ b
f (x) dx = lim
f (x) dx.
b→∞
a
a
(2) If f is continuous on (−∞, b], then
ˆ b
ˆ b
f (x) dx = lim
f (x) dx.
a→−∞
−∞
a
(3) If f is continuous on [a, b) then
ˆ b
ˆ c
f (x) dx = lim
f (x) dx.
−
c→b
a
a
(4) If f is continuous on (a, b] then
ˆ b
ˆ b
f (x) dx = lim
f (x) dx.
+
c→a
a
c
In each case, if the limit exists and is finite we say that the improper integral converges and
the limit is the value of the improper integral. Otherwise the improper integral diverges.
Similarly, if f becomes infinite at an interior point d ∈ [a, b], then
ˆ b
ˆ d
ˆ b
f (x) dx =
f (x) dx +
f (x) dx.
a
a
d
This integral (on [a, b]) converges if both integrals (on [a, d] and on [d, b]) converges. Otherwise,
the integral from a to b diverges.
ˆ
ˆ
∞
a
Finally, if f is continuous on (−∞, ∞) and if
ˆ ∞
f (x) dx converges and its value is
−∞
ˆ
ˆ
∞
−∞
f (x) dx +
−∞
f (x) dx both converge, then
a
ˆ
a
f (x) dx =
−∞
f (x) dx and
∞
f (x) dx.
a
If either one or both of the integrals on the right-hand side of this equation diverge, the integral
diverges.
Example 5.2. In each part, determine whether the improper integral converges or diverges,
and find its value if it converges.
12. 12
INTEGRATION TECHNIQUES
ˆ
(22)
(23)
∞
ˆ1 ∞
ˆ−∞
∞
1
dx.
xp
e−|x| dx.
cos x dx.
(24)
−∞
Exercises 5.3. In each part, determine whether the improper integral converges or diverges,
and find its value if it converges.
ˆ 1
dx
(1)
.
2
−1 x
ˆ 16
dx
√ .
(2)
4
x
ˆ0 1
dx
√
(3)
.
2
ˆ0 ∞ 1 − x
ln x
dx.
(4)
ˆ1 ∞ x
dx
(5)
.
1.001
ˆ1 1 x
dx
(6)
1 .
−8 x 3
ˆ 1
dx
(7)
.
0.999
ˆ0 ∞x
x dx
(8)
3 .
2
−∞ (x + 4) 2
ˆ ∞
16 tan−1 x
dx.
(9)
2
ˆ0 ∞ 1 + x
2e−θ sin θ dθ.
(10)
ˆ0 1
(11)
− ln x dx.
ˆ0 ∞
dx
(12)
.
2
ˆ0 1 (x + 1)(x + 1)
(13)
−x ln |x| dx.
ˆ−1
∞
dx
(14)
.
x + e−x
ˆ−∞ e
∞
(15)
ln(ln x).
ee
A comparison test.
Theorem 5.1 (Direct Comparison Test). Let f and g be continuous on [a, ∞) and suppose
that 0 ≤ f (x) ≤ g(x) for all x ≥ a.
ˆ ∞
ˆ ∞
(i) If
g(x) dx converges then
f (x) dx converges.
a
a
13. INTEGRATION TECHNIQUES
ˆ
(ii) If
ˆ
∞
f (x) dx diverges then
a
13
∞
g(x) dx diverges.
a
Example 5.4. In each part, determine whether the improper integral converges or diverges.
ˆ ∞
1
(1)
dx.
x
ˆ0 ∞ x + e
2
e−x dx.
(2)
ˆ0 ∞
2 + sin x
√
(3)
dx.
x
1
ˆ ∞
sin2 x
(4)
dx.
2
ˆ1 ∞ x
dx
√
.
(5)
4
ˆ1 ∞ x + 5
ln x
√ dx.
(6)
ˆ1 ∞ x
dx
(7)
.
5x + 2x
1
ˆ ∞
dx
(8)
.
x
ˆ0 ∞ 1 + e
1 + sin x
dx.
(9)
x2
ˆπ ∞
dx
(10)
.
ˆ2 1 ln x
ln x
√ dx.
(11)
x
0
14. 14
INTEGRATION TECHNIQUES
Answers
cos(ax)
+ c.
a
x
1
Answers 1.2. tan−1
+ c.
a
a
x5 10x3
−
+ 25x + c.
Answers 1.3.
5
3
x−3
Answers 1.4. sin−1
+ c.
2
Answers 1.1. −
Answers 1.5. ln x2 + 2 x + 5 −
3
tan−1
4
x+1
2
Answers 2.1.
(1) sin x − x cos x + c.
(2) x2 ex − 2xex + 2ex + c.
ex (cos x + sin x)
(3)
+ c.
2
2
2
x ln x x
(4)
−
+ c.
2
4
(5) x ln x − x + c.
√
(6) x sin−1 x + 1 − x2 + c.
2
ex (x2 − 1)
+ c.
(7)
2
(8) x tan x + ln(cos x) + c.
3x
x3x
− 2 + c.
(9)
ln 3 ln 3
x(cos(ln x) + sin(ln x))
(10)
+ c.
2
4 x
3 x
2 x
(11) x e − 4x e + 12x e − 24xex + 24ex + c.
x ln x x
− + c.
(12)
2
√ √x 2 √x
(13) 2 xe − 2e + c.
(14) −x2 cos x + 2x sin x + 2 cos x + c.
xe2x e2x
(15)
−
+ c.
2 −x 4 −x
(16) −xe − e + c.
2e2x cos(3x) 3e2x sin(3x)
(17)
+
+ c.
13
13
1
(18) x tan−1 x − ln(1 + x2 ) + c.
2
2
x4 e−x
2
2
(19) −
− x2 e−x − e−x + c.
2 √
−1
esin x ( 1 − x2 + x)
(20)
+ c.
2
5
3
2
2
(21) (1 − x) 2 − (1 − x) 2 + c.
5
3
15
(22) 4 ln 2 −
+ c.
16
+ c.
15. INTEGRATION TECHNIQUES
Answers 3.1.
(1) −
cos5 x
+ c.
5
cos7 x cos5 x
−
+ c.
7
5
3
3
11
4
2
2
(3) sin 2 x − sin 2 x +
sin 2 x + c.
3
7
11
sin3 x
(4) sin x −
+ c.
3
cos10 x cos8
(5)
−
+ c.
10
8
x sin(2x)
+ c.
(6) −
2
4
3x sin(2x) sin(4x)
(7)
+
+
c.
8
4
32
x sin(4x)
+ c.
(8) −
8
32
sec5 x sec3 x
(9)
−
+ c.
5
3
(10) tan x − x + c.
tan5 x tan3 x
(11)
+
+ c.
5
3
tan2 x
+ ln | cos x| + c.
(12)
2
7
sec x sec5 x
(13)
−
+ c.
7
5
(14) ln | sec x + tan x| + c.
7
3
2
2
(15) tan 2 + tan 2 +c.
7
3
1
3
(16) − cot x + cot x + x + c.
3
(17) − ln | csc x + cot x| + c.
csc3 x csc5 x
−
+ c.
(18)
3
5
√
5
9
4
2
(19) − cos x + cos 2 x − cos 2 x + c.
5
9
√
4 − x2
Answers 3.2.
(1) (a) −
+ c.
4x
√
x + 9 + x2
+ c.
(b) ln
3
√
x
(c) x2 − 25 − 5 sec−1 + c.
5
1
x
−1
(d) tan x +
+ c.
2
2(1 + x2 )
√
x x 4 − x2
(e) 2 sin−1 +
+ c.
2
2
√
(f) √x2 − 1 − sec−1 x + c.
x2 − 2
(g)
+ c.
2x
(2)
15
16. 16
INTEGRATION TECHNIQUES
√
3x
9x2 − 4 − 2 sec−1
+ c.
2
1
x+1
(i) tan−1
+ c.
3
3
√
ex 1 − e2x
1 −1 x
+ c.
(j) sin (e ) +
2
2
(k) ln 1 + sin2 x + sin x + c.
1
2x + 1
2x + 1
(l)
tan−1
+
+ c.
32
2
16(4x2 + 4x + 5)
x−2
(m) ln x2 − 4 x + 8 + 3 tan−1
+ c.
2
5
13
x−2
(n) ln x2 − 4 x + 8 +
tan−1
+ c.
2
2
2
x3
(o) −
3 + c.
3(x2 − 1) 2
π(π + 2)
(2)
.
2
1
1
Answers 4.1.
(1) ln |x − 1| − ln |x + 2| + c.
3
3
(2) 2 ln |x| + 4 ln |x + 1| − 3 ln |x − 1| + c.
3
1
(3) x2 + ln |x − 4| − ln |x + 2| + c.
2
2
9
(4) −
+ 6 ln |x| − ln |x + 1| + c.
x+1
(5) −5 tan−1 x + 2 ln |x| + c.
3
7
x+1
(6) ln x2 + 2 x + 5 − tan−1
+ 2 ln |x + 2| + c.
2
2
2
7
8
(7) ln |x − 3| + ln |x + 2| + c.
5
5
5
1
5
(8) − ln |x − 2| − ln |x| + ln |x − 1| + ln |x − 3| + c.
2
6
3
3
(9) −
+ ln |x − 3| + c.
x−3
1
1
(10) −
+
+ c.
x + 1 (x + 1)2
1
1
1
(11) ln |x + 1| + ln |x − 1| −
+ c.
2
2
x−1
3
1
3
(12) −
ln x2 + 1 − tan−1 x + ln |x − 2| + c.
10
5
5
√
√
5
1
3 2
x 2
2
−1
(13) ln |3 + x| +
− ln x + 2 +
tan
2(x2 + 2) 2
2
2
2
1
(14) x + ln |3 x + 1| + c.
3
9
x2
1
(15) x +
+ ln x2 + 1 + tan−1 x − 2 ln |x − 1| −
+ c.
2
x−1
x2
(16) x +
− ln |x| + 2 ln |x − 1| + c.
2
(h)
+ c.
17. INTEGRATION TECHNIQUES
(17) ln |sec x + tan x| + c.
1
125
x2
− 6 x − ln |x + 1| +
ln |5 + x| + c.
(18)
2
4
4
1
1
1
(19) x + ln |x − 1| − ln |x + 1| − tan−1 x + c.
4
4 √
2
√
1
3
(2 x + 1) 3
1
2
−1
(20) − ln x + x + 1 −
tan
+ ln |x − 1| + c.
6
3
3
3
−x
x
(21) −e + x − ln |e − 1| + c.
1
1
(22) − ln |2 cos x − 1| + ln |1 + cos x| + c.
3
3
1
3
3
1
(23) sin x −
− ln |sin x + 1| + ln |sin x − 1| −
+ c.
4(sin x + 1) 4
4
4(sin x − 1)
√
√
1
1
1
(24) √
− ln 1 − x2 + 1 + ln 1 − x2 − 1 + c.
2
1 − x2 2
1
(25) ln x2 + 2 x + 2 − tan−1 (x + 1) − 2
+ c.
x + 2x + 2
Answers 5.1. See, Answers 5.2, (5).
Answers 5.2.
(1) Convergent, 2.
(2) Divergent.
(3) Convergent, 2.
(4) Divergent.
(5) Divergent.
(6) Convergent, 1.
(7) Divergent.
1
(8) Convergent, .
e
(9) Divergent.
(10) Convergent, 1.
(11) Convergent, 0.
π
(12) Convergent, .
2
√
3
(13) Convergent, 3 + 3 2.
(14) Divergent.
(15) Divergent if p ≥ 1. Convergent if p < 1,
(16)
(17)
(18)
(19)
(20)
(21)
1
.
1−p
Divergent.
Convergent, 1.
Convergent, π.
Convergent, 0.
Divergent.
Divergent.
(22) Divergent if p ≤ 1. Convergent if p > 1,
(23) Convergent, 2.
(24) Divergent.
Answers 5.3.
(1) Divergent.
1
.
p−1
17