Mathematical Induction
CMSC 56 | Discrete Mathematical Structure for Computer Science
October 18, 2018
Instructor: Allyn Joy D. Calcaben
College of Arts & Sciences
University of the Philippines Visayas
Mathematical Induction
CMSC 56 | Discrete Mathematical Structure for Computer Science
October 18, 2018
Instructor: Allyn Joy D. Calcaben
College of Arts & Sciences
University of the Philippines Visayas
Finally, I was able to put together the talk about skip list... I am still not liking my explanation of the scan-forward part to be bounded by a geometric random variable... However... enjoy
Integrals by Trigonometric SubstitutionPablo Antuna
In this video we learn the basics of trigonometric substitution and why it works. We talk about all the basic cases of integrals you can solve using trigonometric substitution.
For more lessons and presentations:
Finally, I was able to put together the talk about skip list... I am still not liking my explanation of the scan-forward part to be bounded by a geometric random variable... However... enjoy
Integrals by Trigonometric SubstitutionPablo Antuna
In this video we learn the basics of trigonometric substitution and why it works. We talk about all the basic cases of integrals you can solve using trigonometric substitution.
For more lessons and presentations:
Welcome to TechSoup New Member Orientation and Q&A (May 2024).pdfTechSoup
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Students, digital devices and success - Andreas Schleicher - 27 May 2024..pptxEduSkills OECD
Andreas Schleicher presents at the OECD webinar ‘Digital devices in schools: detrimental distraction or secret to success?’ on 27 May 2024. The presentation was based on findings from PISA 2022 results and the webinar helped launch the PISA in Focus ‘Managing screen time: How to protect and equip students against distraction’ https://www.oecd-ilibrary.org/education/managing-screen-time_7c225af4-en and the OECD Education Policy Perspective ‘Students, digital devices and success’ can be found here - https://oe.cd/il/5yV
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.
The Indian economy is classified into different sectors to simplify the analysis and understanding of economic activities. For Class 10, it's essential to grasp the sectors of the Indian economy, understand their characteristics, and recognize their importance. This guide will provide detailed notes on the Sectors of the Indian Economy Class 10, using specific long-tail keywords to enhance comprehension.
For more information, visit-www.vavaclasses.com
2024.06.01 Introducing a competency framework for languag learning materials ...Sandy Millin
http://sandymillin.wordpress.com/iateflwebinar2024
Published classroom materials form the basis of syllabuses, drive teacher professional development, and have a potentially huge influence on learners, teachers and education systems. All teachers also create their own materials, whether a few sentences on a blackboard, a highly-structured fully-realised online course, or anything in between. Despite this, the knowledge and skills needed to create effective language learning materials are rarely part of teacher training, and are mostly learnt by trial and error.
Knowledge and skills frameworks, generally called competency frameworks, for ELT teachers, trainers and managers have existed for a few years now. However, until I created one for my MA dissertation, there wasn’t one drawing together what we need to know and do to be able to effectively produce language learning materials.
This webinar will introduce you to my framework, highlighting the key competencies I identified from my research. It will also show how anybody involved in language teaching (any language, not just English!), teacher training, managing schools or developing language learning materials can benefit from using the framework.
Instructions for Submissions thorugh G- Classroom.pptxJheel Barad
This presentation provides a briefing on how to upload submissions and documents in Google Classroom. It was prepared as part of an orientation for new Sainik School in-service teacher trainees. As a training officer, my goal is to ensure that you are comfortable and proficient with this essential tool for managing assignments and fostering student engagement.
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.
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.
1. Table of Integrals*
Integrals with Logarithms
Basic Forms
1
x dx =
xn+1
n+1
1
dx = ln |x|
x
(2)
x
dx =
a+x
(1)
n
udv = uv −
vdu
√
x ax + bdx =
x(a + x) − a ln
√
x+
√
x+a
(25)
ln axdx = x ln ax − x
(3)
1
1
dx = ln |ax + b|
ax + b
a
ln ax
1
dx = (ln ax)2
x
2
√
2
(−2b2 + abx + 3a2 x2 ) ax + b (26)
15a2
(42)
(43)
ln(ax + b)dx =
1
x(ax + b)dx = 3/2 (2ax + b) ax(ax + b)
4a
√
(27)
−b2 ln a x + a(ax + b)
(4)
x+
b
a
ln(ax + b) − x, a = 0
ln(x2 + a2 ) dx = x ln(x2 + a2 ) + 2a tan−1
Integrals of Rational Functions
1
1
dx = −
(x + a)2
x+a
(x + a)n dx =
b
b2
x
− 2 +
x3 (ax + b)
12a
8a x
3
√
b3
+ 5/2 ln a x + a(ax + b)
(28)
8a
x3 (ax + b)dx =
(5)
(x + a)n+1
, n = −1
n+1
(6)
n+1
x(x + a)n dx =
(x + a)
((n + 1)x − a)
(n + 1)(n + 2)
1
x2 ± a2 dx = x
2
(7)
1
x2 ± a2 ± a2 ln x +
2
ln(x2 − a2 ) dx = x ln(x2 − a2 ) + a ln
ln ax2 + bx + c dx =
x2 ± a2
(9)
x
1
dx = ln |a2 + x2 |
a2 + x2
2
(10)
a2 − x2 dx =
x
2
a2
x
x
dx = x − a tan−1
+ x2
a
x3
1
1
dx = x2 − a2 ln |a2 + x2 |
2 + x2
a
2
2
1
2
2ax + b
dx = √
tan−1 √
ax2 + bx + c
4ac − b2
4ac − b2
1
1
a+x
dx =
ln
, a=b
(x + a)(x + b)
b−a
b+x
x
a
dx =
+ ln |a + x|
(x + a)2
a+x
ax2
x
1
dx =
ln |ax2 + bx + c|
+ bx + c
2a
2ax + b
b
− √
tan−1 √
a 4ac − b2
4ac − b2
(11)
(13)
x
√
dx =
x2 ± a2
(15)
√
(ax + b)3/2 dx =
√
(16)
(19)
*
«
x
dx = −
a−x
ax2 + bx + cdx =
(33)
eax dx =
(34)
√
a2 − x2
b + 2ax
4a
4ac − b2
ln 2ax + b + 2
8a3/2
xeax dx =
x
1
− 2
a
a
(37)
(24)
eax
(53)
x2
2x
2
− 2 + 3
a
a
a
(54)
eax
(55)
x3 ex dx = x3 − 3x2 + 6x − 6 ex
ax2
x2 eax dx =
(51)
(52)
x2 ex dx = x2 − 2x + 2 ex
(56)
+ bx + c
xn eax dx =
2
ax2 + bx + c
1
1
√
dx = √ ln 2ax + b + 2
a
ax2 + bx + c
xn eax
n
−
a
a
(38)
a(ax2 + bx + c)
dx
x
= √
+ x2 )3/2
a2 a2 + x2
(57)
(58)
ta−1 e−t dt
x
√
√
2
i π
eax dx = − √ erf ix a
2 a
√
√
2
π
e−ax dx = √ erf x a
2 a
2
2
x2 e−ax dx =
1
4
(59)
(60)
1 −ax2
e
2a
(61)
√
π
x −ax2
erf(x a) −
e
a3
2a
(62)
xe−ax dx = −
(40)
(41)
xn−1 eax dx
(−1)n
Γ[1 + n, −ax],
an+1
where Γ(a, x) =
x
1
√
dx =
ax2 + bx + c
a
ax2 + bx + c
b
− 3/2 ln 2ax + b + 2 a(ax2 + bx + c)
2a
(a2
(50)
xex dx = (x − 1)ex
x2 ± a2
(39)
(23)
x(a − x)
x−a
1 ax
e
a
√
√
1 √ ax
i π
xe + 3/2 erf i ax ,
a
2a
x
2
2
where erf(x) = √
e−t dt
π 0
(35)
a(ax2 + bx+ c)
× −3b + 2abx + 8a(c + ax )
√
+3(b3 − 4abc) ln b + 2ax + 2 a
(49)
xeax dx =
ax2 + bx + c
√
1
ax2 + bx + c =
2 a
48a5/2
ln a2 − b2 x2
Integrals with Exponentials
1 2
a ln x +
2
x2 ± a2
(48)
(32)
(21)
√
2a) x ± a
x(a − x) − a tan
x2 ± a2
∞
(22)
−1
(31)
xn eax dx =
(20)
(47)
ln(ax + b)
1
x ln a2 − b2 x2 dx = − x2 +
2
1
a2
x2 − 2
2
b
(36)
2
(ax + b)5/2
5a
2
x
dx = (x
3
x±a
x2
1
dx = x
2
± a2
x2
2
√
ax + b
2b
2x
+
3a
3
3/2
x2 ± a2
x
√
dx = −
a2 − x2
(14)
x
√
1
dx = −2 a − x
a−x
ax + bdx =
1 2
x2 ± a2 dx =
x ± a2
3
(17)
√
2
2
x x − adx = a(x − a)3/2 + (x − a)5/2
3
5
√
1 2
x
a tan−1 √
2
a2 − x2
(30)
a2 − x2 +
1
x
√
dx = sin−1
a
a2 − x2
(18)
2ax + b
4ac − b2 tan−1 √
4ac − b2
bx
1
− x2
2a
4
1
b2
+
x2 − 2
2
a
(12)
+
√
1
√
dx = 2 x ± a
x±a
√
1
x
2
1
√
dx = ln x +
x2 ± a2
Integrals with Roots
√
2
x − adx = (x − a)3/2
3
x+a
− 2x (46)
x−a
x ln(ax + b)dx =
(8)
1
1
x
dx = tan−1
a2 + x2
a
a
x
− 2x (45)
a
b
+ x ln ax2 + bx + c
2a
− 2x +
(29)
1
dx = tan−1 x
1 + x2
1
a
(44)
2013. From http://integral-table.com, last revised July 19, 2013. This material is provided as is without warranty or representation about the accuracy, correctness or
suitability of this material for any purpose. This work is licensed under the Creative Commons Attribution-Noncommercial-ShareAlike 3.0 United States License. To view a copy
of this license, visit http://creativecommons.org/licenses/by-nc-sa/3.0/ or send a letter to Creative Commons, 171 Second Street, Suite 300, San Francisco, California, 94105,
USA.
2. Integrals with Trigonometric Functions
sec3 x dx =
1
sin axdx = − cos ax
a
sin2 axdx =
1
1
sec x tan x + ln | sec x + tan x|
2
2
sec x tan xdx = sec x
x
sin 2ax
−
2
4a
sin axdx =
1
sec x tan xdx = secn x, n = 0
n
1 1−n 3
,
, , cos2 ax
2
2
2
(66)
1
sin ax
a
cos ax sin bxdx =
(68)
1
1
csc3 xdx = − cot x csc x + ln | csc x − cot x|
2
2
1
csc x cot xdx = − cscn x, n = 0
n
n
sec x csc xdx = ln | tan x|
(69)
(70)
(107)
1 x
e (cos x − x cos x + x sin x)
2
(108)
xex cos xdx =
1 x
e (x cos x − sin x + x sin x)
2
(109)
(86)
(87)
(88)
(89)
x cos xdx = cos x + x sin x
1
x
cos ax + sin ax
a2
a
x cos axdx =
sin[(2a − b)x]
sin ax cos bxdx = −
4(2a − b)
sin[(2a + b)x]
sin bx
−
+
2b
4(2a + b)
(91)
(92)
(93)
(94)
(95)
(72)
2x cos ax
a2 x2 − 2
x cos axdx =
+
sin ax
2
a
a3
2
1
sin3 x
3
(73)
cos[(2a − b)x]
cos bx
cos ax sin bxdx =
−
4(2a − b)
2b
cos[(2a + b)x]
−
4(2a + b)
1
xn cosxdx = − (i)n+1 [Γ(n + 1, −ix)
2
+(−1)n Γ(n + 1, ix)]
2
1
cos3 ax
3a
eax cosh bxdx =
ax
e
2
[a cosh bx − b sinh bx]
a − b2
2ax
x
e
+
4a
2
sinh axdx =
2
x2 cos xdx = 2x cos x + x2 − 2 sin x
(96)
(97)
eax tanh bxdx =
(a+2b)x
e
a
a
2bx
(a + 2b) 2 F1 1 + 2b , 1, 2 + 2b , −e
a
1
, 1, 1E, −e2bx
− eax 2 F1
a −1 ax 2b
ax
e − 2 tan [e ]
a
tanh ax dx =
sin[2(a − b)x]
x
sin 2ax
−
−
4
8a
16(a − b)
sin[2(a + b)x]
sin 2bx
+
−
(76)
8b
16(a + b)
1
(ia)1−n [(−1)n Γ(n + 1, −iax)
2
−Γ(n + 1, ixa)]
(98)
1
tan ax
a
(79)
(80)
1
1
ln cos ax +
sec2 ax
a
2a
sec xdx = ln | sec x + tan x| = 2 tanh−1 tan
1
tan ax
a
a=b
(113)
a=b
a = b (114)
a=b
1
ln cosh ax
a
(115)
1
[a sin ax cosh bx
a2 + b2
+b cos ax sinh bx]
(116)
cos ax cosh bxdx =
(100)
1
[b cos ax cosh bx+
a2 + b2
a sin ax sinh bx]
(117)
(101)
cos ax sinh bxdx =
(81)
x2 sin axdx =
2 2
1
[−a cos ax cosh bx+
a2 + b2
b sin ax sinh bx]
(118)
2−a x
2x sin ax
cos ax +
a3
a2
(102)
1
xn sin xdx = − (i)n [Γ(n + 1, −ix) − (−1)n Γ(n + 1, −ix)]
2
(103)
1
[b cosh bx sin ax−
a2 + b2
a cos ax sinh bx]
(119)
sin ax sinh bxdx =
Products of Trigonometric Functions and
Exponentials
sinh ax cosh axdx =
x
2
(82)
ex sin xdx =
1 x
e (sin x − cos x)
2
(83)
ebx sin axdx =
1
ebx (b sin ax − a cos ax)
a2 + b2
1
[−2ax + sinh 2ax]
4a
(120)
(104)
1
[b cosh bx sinh ax
b2 − a2
−a cosh ax sinh bx]
(121)
sinh ax cosh bxdx =
sec2 axdx =
(112)
sin ax cosh bxdx =
tann+1 ax
×
a(1 + n)
n+3
n+1
, 1,
, − tan2 ax
2
2
tan3 axdx =
sin ax
x cos ax
+
a
a2
x sin axdx = −
(78)
tann axdx =
2 F1
(111)
a=b
(99)
(77)
1
tan axdx = − ln cos ax
a
tan2 axdx = −x +
x sin xdx = −x cos x + sin x
x2 sin xdx = 2 − x2 cos x + 2x sin x
sin 4ax
x
−
8
32a
a=b
xn cosaxdx =
sin2 ax cos2 bxdx =
sin2 ax cos2 axdx =
(110)
1
cosh ax
a
eax sinh bxdx =
ax
e
2
[−b cosh bx + a sinh bx]
a − b2
2ax
x
e
−
4a
2
(74)
(75)
1
sinh ax
a
(90)
Products of Trigonometric Functions and
Monomials
cos[(a − b)x]
cos[(a + b)x]
−
,a = b
2(a − b)
2(a + b)
(71)
cos2 ax sin axdx = −
1
ebx (a sin ax + b cos ax)
a2 + b2
cosh axdx =
3 sin ax
sin 3ax
+
4a
12a
sin2 x cos xdx =
ebx cos axdx =
Integrals of Hyperbolic Functions
1
csc axdx = − cot ax
a
(67)
1
cos1+p ax×
a(1 + p)
1+p 1 3+p
, ,
, cos2 ax
2
2
2
cos3 axdx =
x
= ln | csc x − cot x| + C
csc xdx = ln tan
2
2
sin 2ax
x
cos2 axdx = +
2
4a
2 F1
(106)
xex sin xdx =
(85)
(65)
3 cos ax
cos 3ax
+
4a
12a
cos axdx =
cosp axdx = −
1
sec2 x
2
n
sin3 axdx = −
1 x
e (sin x + cos x)
2
(64)
n
2 F1
ex cos xdx =
(63)
sec2 x tan xdx =
1
− cos ax
a
(84)
(105)