Microstrip Antenna Design For Ultra-Wide Band Applicationsinventionjournals
1) The document describes the design of a microstrip patch antenna for ultra-wideband applications. Slots are added to the patch and ground plane to enhance the bandwidth.
2) Simulation results show the antenna achieves an impedance bandwidth of 2.2-5.6 GHz with return loss less than -10 dB. The gain is up to 6.5 dB and radiation patterns are dipole-like in the E-plane and omnidirectional in the H-plane.
3) The antenna design achieves good performance across the UWB frequency range and could be suitable for applications such as wireless monitors and printers.
This document discusses a home automation project using Internet of Things (IoT) technology. The project aims to automate household activities and appliances like lighting, HVAC, entertainment, and security using wireless connectivity. Key components include ESP8266 WiFi modules, transistors, relays, and other basic electronic components. The system is designed to improve convenience, comfort, energy efficiency, and security for homeowners while keeping the system affordable and simple to use. Quality assurance methods like failure testing and statistical control are applied to ensure requirements are fulfilled.
SMART HOME AUTOMATION USING MOBILE APPLICATIONEklavya Sharma
This document outlines a smart home automation project using a mobile application to control home appliances via an Arduino board and Bluetooth module. The objectives are to understand smart home concepts, establish serial communication between Arduino and a mobile device, and design a user interface. The system allows controlling AC loads from an Android phone app through an Arduino, Bluetooth module, relays, and relay driver IC. A block diagram and flowchart illustrate the components and process. An App Inventor mobile app is created to provide the human-machine interface.
These slides give an extensive knowledge about the photo diode. It covers the circuit diagram and its energy band diagram. Also includes important information about noise factors and resoponsitivities.
Access the video from this presentation for free from
http://www.rohde-schwarz-usa.com/DebuggingEMISS_On-Demand.html
Overview:
Electromagnetic interference is increasingly becoming a problem in complex systems that must interoperate in both digital and RF domains. When failures due to EMI occur it is often difficult to track down the sources of such failures using standard test receivers and spectrum analyzers. The unique ability of real-time spectrum analysis and synchronous time domain signal acquisition to capture transient events can quickly reveals details about the sources of EMI.
What You Will Learn:
How to isolate and analyze sources of EMI using an oscilloscope
Measurement considerations for correlating time and frequency domains
Near field probing basics
Presented By:
Dave Rishavy, Product Manager Oscilloscopes, Rohde & Schwarz
Dave Rishavy has a BS in Electrical Engineering from Florida State University and an MBA from the University of Colorado. Prior to joining Rohde and Schwarz, Mr. Rishavy gained over 15 years of experience in the test and measurement field at Agilent Technologies. This included positions in a wide range of technical marketing areas such as application engineering, product marketing, marketing management and strategic product planning. While at Agilent, Dave led the marketing and industry segment teams for the Infiniium line of oscilloscopes as well as high end logic analysis.
En telecomunicaciones, 5G son las siglas utilizadas para referirse a la quinta generación de tecnologías de telefonía móvil. Es la sucesora de la tecnología 4G la cual le provee conectividad a la mayoría de teléfonos móviles actuales
1) The document describes an Android-based home automation project that uses Bluetooth to control appliances. A microcontroller and Bluetooth module are used to connect household devices like lights and fans to an Android mobile app for remote control.
2) The system allows users to control appliances from their phone, providing assistance for physically challenged individuals. Components include an ATmega8 microcontroller, HC-05 Bluetooth module, relays, LEDs, and Android mobile phone.
3) The app interface provides on/off buttons for devices. Advantages are wireless control and assistance for disabled people. Limitations include small Bluetooth range and higher cost compared to traditional remote controls. The technology could help monitor weather, temperatures, and power appliances like
The document provides information on microwave engineering and transmission lines. It discusses topics such as the frequency range of microwave engineering (1 GHz and above), transmission line theory, characteristic impedance, propagation velocity, lossless transmission lines, and formulas for calculating the characteristic impedance of different transmission line structures including coaxial cable, striplines, and dual striplines.
Microstrip Antenna Design For Ultra-Wide Band Applicationsinventionjournals
1) The document describes the design of a microstrip patch antenna for ultra-wideband applications. Slots are added to the patch and ground plane to enhance the bandwidth.
2) Simulation results show the antenna achieves an impedance bandwidth of 2.2-5.6 GHz with return loss less than -10 dB. The gain is up to 6.5 dB and radiation patterns are dipole-like in the E-plane and omnidirectional in the H-plane.
3) The antenna design achieves good performance across the UWB frequency range and could be suitable for applications such as wireless monitors and printers.
This document discusses a home automation project using Internet of Things (IoT) technology. The project aims to automate household activities and appliances like lighting, HVAC, entertainment, and security using wireless connectivity. Key components include ESP8266 WiFi modules, transistors, relays, and other basic electronic components. The system is designed to improve convenience, comfort, energy efficiency, and security for homeowners while keeping the system affordable and simple to use. Quality assurance methods like failure testing and statistical control are applied to ensure requirements are fulfilled.
SMART HOME AUTOMATION USING MOBILE APPLICATIONEklavya Sharma
This document outlines a smart home automation project using a mobile application to control home appliances via an Arduino board and Bluetooth module. The objectives are to understand smart home concepts, establish serial communication between Arduino and a mobile device, and design a user interface. The system allows controlling AC loads from an Android phone app through an Arduino, Bluetooth module, relays, and relay driver IC. A block diagram and flowchart illustrate the components and process. An App Inventor mobile app is created to provide the human-machine interface.
These slides give an extensive knowledge about the photo diode. It covers the circuit diagram and its energy band diagram. Also includes important information about noise factors and resoponsitivities.
Access the video from this presentation for free from
http://www.rohde-schwarz-usa.com/DebuggingEMISS_On-Demand.html
Overview:
Electromagnetic interference is increasingly becoming a problem in complex systems that must interoperate in both digital and RF domains. When failures due to EMI occur it is often difficult to track down the sources of such failures using standard test receivers and spectrum analyzers. The unique ability of real-time spectrum analysis and synchronous time domain signal acquisition to capture transient events can quickly reveals details about the sources of EMI.
What You Will Learn:
How to isolate and analyze sources of EMI using an oscilloscope
Measurement considerations for correlating time and frequency domains
Near field probing basics
Presented By:
Dave Rishavy, Product Manager Oscilloscopes, Rohde & Schwarz
Dave Rishavy has a BS in Electrical Engineering from Florida State University and an MBA from the University of Colorado. Prior to joining Rohde and Schwarz, Mr. Rishavy gained over 15 years of experience in the test and measurement field at Agilent Technologies. This included positions in a wide range of technical marketing areas such as application engineering, product marketing, marketing management and strategic product planning. While at Agilent, Dave led the marketing and industry segment teams for the Infiniium line of oscilloscopes as well as high end logic analysis.
En telecomunicaciones, 5G son las siglas utilizadas para referirse a la quinta generación de tecnologías de telefonía móvil. Es la sucesora de la tecnología 4G la cual le provee conectividad a la mayoría de teléfonos móviles actuales
1) The document describes an Android-based home automation project that uses Bluetooth to control appliances. A microcontroller and Bluetooth module are used to connect household devices like lights and fans to an Android mobile app for remote control.
2) The system allows users to control appliances from their phone, providing assistance for physically challenged individuals. Components include an ATmega8 microcontroller, HC-05 Bluetooth module, relays, LEDs, and Android mobile phone.
3) The app interface provides on/off buttons for devices. Advantages are wireless control and assistance for disabled people. Limitations include small Bluetooth range and higher cost compared to traditional remote controls. The technology could help monitor weather, temperatures, and power appliances like
The document provides information on microwave engineering and transmission lines. It discusses topics such as the frequency range of microwave engineering (1 GHz and above), transmission line theory, characteristic impedance, propagation velocity, lossless transmission lines, and formulas for calculating the characteristic impedance of different transmission line structures including coaxial cable, striplines, and dual striplines.
This document provides an overview of microstrip patch antennas, also known as patch antennas. It defines patch antennas as consisting of a metal patch on top of a grounded dielectric substrate, which are useful at microwave frequencies above 1 GHz. The document discusses the geometry, advantages, disadvantages, feeding techniques, basic properties including resonance frequency and bandwidth, radiation pattern, and applications of microstrip patch antennas. The main applications mentioned are in mobiles, satellites, GPS, WiMAX, medical devices, and radar.
The document discusses 5G and 6G mobile technologies. It provides an overview of the evolution from 1G to 5G networks, describing some key 5G technologies like millimeter wave, small cells, massive MIMO, and beamforming. It then introduces 6G, explaining that 6G networks are expected to utilize terahertz bands and technologies like AI, optical wireless communication, and 3D networking. Some advantages of 6G mentioned include extremely high speeds, low latency, improved security and personalization, and enabling new applications like connected robotics.
The document discusses enabling technologies for ultra-reliable low-latency communication (URLLC) and their applications. It examines URLLC from a systems perspective, covering modeling approaches like extreme value theory and tools from mathematical finance. The document also explores URLLC use cases such as vehicle-to-everything communication, virtual reality, and wireless edge machine learning.
This document describes a home automation system that allows users to control devices in their home from their mobile phone using Bluetooth. The system connects an Android phone to a microcontroller using a Bluetooth module. The user can then control electrical devices and appliances connected to relays through the microcontroller. This provides convenience and accessibility from anywhere. It also improves energy efficiency and security for homes.
Fiber optics use total internal reflection to transmit light signals for communication. They convert electrical signals to optical signals using a transmitter, transmit the light through the optical fiber, then convert it back to an electrical signal using a receiver. Fiber optics have huge bandwidth potential and are immune to electromagnetic interference but were initially more expensive than electrical cables. Key advantages include bandwidths over 250Gbps, low signal loss, and resistance to corrosion.
This document provides an overview of optical DWDM fundamentals, including:
- Key terminology used in optical networks such as decibels, wavelength, frequency, and fiber impairments.
- Characteristics of optical fiber including different fiber types, fiber dimensions, and how light propagates through total internal reflection.
- Factors that reduce optical power over distance, specifically attenuation from absorption and scattering in the fiber material.
This document provides an outline and overview of topics related to mobile radio propagation, including:
- Definitions of key terms like wavelength, frequency, propagation mechanisms, and radio frequency bands.
- Descriptions of different propagation effects like reflection, diffraction, and scattering.
- Explanations of path loss in various environments and how it relates to distance and frequency.
- Discussions of slow fading/shadowing and fast fading/multipath effects on received signals.
- Concepts of Doppler shift, delay spread, and intersymbol interference in mobile radio systems.
This document provides an overview of a project to create a home automation system using GSM technology. The system would allow users to control home appliances via SMS from their mobile phone. It discusses the motivation, objectives, basic working, components, costs, timeline, and pros and cons of the system. The objectives are to enable remote control of appliances without being physically present and reduce power and time wastage. The estimated cost is 3,814 Rs. and the timeline outlines tasks from feasibility to testing over 6 months.
How will sidelink bring a new level of 5G versatility.pdfQualcomm Research
Today, the 5G system mainly operates on a network-to-device communication model, exemplified by enhanced mobile broadband use cases where all data transmissions are between the network (i.e., base station) and devices (e.g., smartphone). However, to fully deliver on the original 5G vision of supporting diverse devices, services, and deployment scenarios, we need to expand the 5G topology further to reach new levels of performance and efficiency.
That is why sidelink communication was introduced in 3GPP standards, designed to facilitate direct communication between devices, independent of connectivity via the cellular infrastructure. Beyond automotive communication, it also benefits many other 5G use cases such as IoT, mobile broadband, and public safety.
Fiber optics use thin strands of glass called optical fibers to transmit light signals over long distances. Light travels through the core of the fiber, which is surrounded by cladding that reflects the light down the length of the fiber. Fiber optic systems include a transmitter that produces light signals, the optical fiber that carries the signals, and a receiver that interprets the signals. Fiber optics have advantages over metal wires like lower costs, higher data capacity, and less signal degradation over long distances.
A smart antenna consists of an antenna array that can change its pattern in response to the signal environment to improve communication system performance. There are two main types: switched beam arrays, which use multiple fixed beams with one beam activated towards the desired signal; and adaptive arrays, which use algorithms to minimize interfering signals and maximize signal-to-interference ratio (SIR). The goals of a smart antenna system are to improve signal gain, interfere rejection, and power efficiency.
Rayleigh Fading Channel In Mobile Digital Communication SystemOUM SAOKOSAL
The document discusses Rayleigh fading channels in mobile digital communication systems. It describes how multipath propagation can cause multipath fading or scintillation. It distinguishes between large-scale fading and small-scale fading. Large-scale fading refers to mean signal attenuation over large areas and variations around the mean due to shadowing. Small-scale fading is also called Rayleigh fading and refers to time spreading of signals and time variance of channels due to small changes in position.
Millimeter wave 5G antennas for smartphonesPei-Che Chang
This document describes research on millimeter-wave antennas for 5G smartphones. It discusses several antenna designs for both 60 GHz and 28 GHz applications. For 60 GHz, a 2012 design integrated a 16-element phased array directly into a printed circuit board. Later designs in 2013 and 2017 explored integrating antenna arrays with reconfigurable polarization into mobile device chassis. A 2014 design proposed a 28 GHz mesh-grid patch antenna array for 5G cellular devices, demonstrating an 11 dBi gain array integrated into a Samsung phone. The document outlines various antenna designs, simulation and measurement results to enable millimeter-wave smartphone connectivity.
This document discusses the GSM wireless telecommunication system. It provides an overview of GSM, including its history and standards body. Key aspects covered include the system architecture with components like the mobile station, base station, mobile switching center, and location registers. Services offered by GSM like basic bearer services, teleservices, and supplementary services are outlined. Performance characteristics, disadvantages, and the radio subsystem, network switching subsystem, and operations subsystem are also summarized. Diagrams illustrate worldwide subscriber growth and the GSM system architecture.
This document discusses monopole antennas. It begins by explaining that a monopole antenna is half of a dipole antenna mounted above a ground plane. Using image theory, the fields of a monopole antenna above a ground plane are equivalent to a dipole antenna of twice the length in free space. The impedance of a monopole antenna is half that of a dipole antenna of the same length. Effects of a finite ground plane are also discussed, noting that the radiation pattern becomes skewed away from the horizontal plane for small ground planes.
The RBS 6401 is a new small cell base station from Ericsson that provides an easy solution for deploying small cells in heterogeneous networks. It has a compact integrated antenna design that allows for flexible placement. The deployment and provisioning process is simplified through features such as self-organizing networks and easy integration with OSS systems. The RBS 6401 supports both LTE and WCDMA as well as WiFi, and provides high capacity through features such as dual carriers and support for a large number of scheduled users.
Descripción de los conceptos fundamentales de redes ópticas pasivas (PON): elementos de la red de acceso óptica, topologías, variantes FTTx, protocolos (TDM y TDMA), tecnologías y estándares (BPON; GPON y EPON) y beneficios.
This document provides an overview of microstrip patch antennas, also known as patch antennas. It defines patch antennas as consisting of a metal patch on top of a grounded dielectric substrate, which are useful at microwave frequencies above 1 GHz. The document discusses the geometry, advantages, disadvantages, feeding techniques, basic properties including resonance frequency and bandwidth, radiation pattern, and applications of microstrip patch antennas. The main applications mentioned are in mobiles, satellites, GPS, WiMAX, medical devices, and radar.
The document discusses 5G and 6G mobile technologies. It provides an overview of the evolution from 1G to 5G networks, describing some key 5G technologies like millimeter wave, small cells, massive MIMO, and beamforming. It then introduces 6G, explaining that 6G networks are expected to utilize terahertz bands and technologies like AI, optical wireless communication, and 3D networking. Some advantages of 6G mentioned include extremely high speeds, low latency, improved security and personalization, and enabling new applications like connected robotics.
The document discusses enabling technologies for ultra-reliable low-latency communication (URLLC) and their applications. It examines URLLC from a systems perspective, covering modeling approaches like extreme value theory and tools from mathematical finance. The document also explores URLLC use cases such as vehicle-to-everything communication, virtual reality, and wireless edge machine learning.
This document describes a home automation system that allows users to control devices in their home from their mobile phone using Bluetooth. The system connects an Android phone to a microcontroller using a Bluetooth module. The user can then control electrical devices and appliances connected to relays through the microcontroller. This provides convenience and accessibility from anywhere. It also improves energy efficiency and security for homes.
Fiber optics use total internal reflection to transmit light signals for communication. They convert electrical signals to optical signals using a transmitter, transmit the light through the optical fiber, then convert it back to an electrical signal using a receiver. Fiber optics have huge bandwidth potential and are immune to electromagnetic interference but were initially more expensive than electrical cables. Key advantages include bandwidths over 250Gbps, low signal loss, and resistance to corrosion.
This document provides an overview of optical DWDM fundamentals, including:
- Key terminology used in optical networks such as decibels, wavelength, frequency, and fiber impairments.
- Characteristics of optical fiber including different fiber types, fiber dimensions, and how light propagates through total internal reflection.
- Factors that reduce optical power over distance, specifically attenuation from absorption and scattering in the fiber material.
This document provides an outline and overview of topics related to mobile radio propagation, including:
- Definitions of key terms like wavelength, frequency, propagation mechanisms, and radio frequency bands.
- Descriptions of different propagation effects like reflection, diffraction, and scattering.
- Explanations of path loss in various environments and how it relates to distance and frequency.
- Discussions of slow fading/shadowing and fast fading/multipath effects on received signals.
- Concepts of Doppler shift, delay spread, and intersymbol interference in mobile radio systems.
This document provides an overview of a project to create a home automation system using GSM technology. The system would allow users to control home appliances via SMS from their mobile phone. It discusses the motivation, objectives, basic working, components, costs, timeline, and pros and cons of the system. The objectives are to enable remote control of appliances without being physically present and reduce power and time wastage. The estimated cost is 3,814 Rs. and the timeline outlines tasks from feasibility to testing over 6 months.
How will sidelink bring a new level of 5G versatility.pdfQualcomm Research
Today, the 5G system mainly operates on a network-to-device communication model, exemplified by enhanced mobile broadband use cases where all data transmissions are between the network (i.e., base station) and devices (e.g., smartphone). However, to fully deliver on the original 5G vision of supporting diverse devices, services, and deployment scenarios, we need to expand the 5G topology further to reach new levels of performance and efficiency.
That is why sidelink communication was introduced in 3GPP standards, designed to facilitate direct communication between devices, independent of connectivity via the cellular infrastructure. Beyond automotive communication, it also benefits many other 5G use cases such as IoT, mobile broadband, and public safety.
Fiber optics use thin strands of glass called optical fibers to transmit light signals over long distances. Light travels through the core of the fiber, which is surrounded by cladding that reflects the light down the length of the fiber. Fiber optic systems include a transmitter that produces light signals, the optical fiber that carries the signals, and a receiver that interprets the signals. Fiber optics have advantages over metal wires like lower costs, higher data capacity, and less signal degradation over long distances.
A smart antenna consists of an antenna array that can change its pattern in response to the signal environment to improve communication system performance. There are two main types: switched beam arrays, which use multiple fixed beams with one beam activated towards the desired signal; and adaptive arrays, which use algorithms to minimize interfering signals and maximize signal-to-interference ratio (SIR). The goals of a smart antenna system are to improve signal gain, interfere rejection, and power efficiency.
Rayleigh Fading Channel In Mobile Digital Communication SystemOUM SAOKOSAL
The document discusses Rayleigh fading channels in mobile digital communication systems. It describes how multipath propagation can cause multipath fading or scintillation. It distinguishes between large-scale fading and small-scale fading. Large-scale fading refers to mean signal attenuation over large areas and variations around the mean due to shadowing. Small-scale fading is also called Rayleigh fading and refers to time spreading of signals and time variance of channels due to small changes in position.
Millimeter wave 5G antennas for smartphonesPei-Che Chang
This document describes research on millimeter-wave antennas for 5G smartphones. It discusses several antenna designs for both 60 GHz and 28 GHz applications. For 60 GHz, a 2012 design integrated a 16-element phased array directly into a printed circuit board. Later designs in 2013 and 2017 explored integrating antenna arrays with reconfigurable polarization into mobile device chassis. A 2014 design proposed a 28 GHz mesh-grid patch antenna array for 5G cellular devices, demonstrating an 11 dBi gain array integrated into a Samsung phone. The document outlines various antenna designs, simulation and measurement results to enable millimeter-wave smartphone connectivity.
This document discusses the GSM wireless telecommunication system. It provides an overview of GSM, including its history and standards body. Key aspects covered include the system architecture with components like the mobile station, base station, mobile switching center, and location registers. Services offered by GSM like basic bearer services, teleservices, and supplementary services are outlined. Performance characteristics, disadvantages, and the radio subsystem, network switching subsystem, and operations subsystem are also summarized. Diagrams illustrate worldwide subscriber growth and the GSM system architecture.
This document discusses monopole antennas. It begins by explaining that a monopole antenna is half of a dipole antenna mounted above a ground plane. Using image theory, the fields of a monopole antenna above a ground plane are equivalent to a dipole antenna of twice the length in free space. The impedance of a monopole antenna is half that of a dipole antenna of the same length. Effects of a finite ground plane are also discussed, noting that the radiation pattern becomes skewed away from the horizontal plane for small ground planes.
The RBS 6401 is a new small cell base station from Ericsson that provides an easy solution for deploying small cells in heterogeneous networks. It has a compact integrated antenna design that allows for flexible placement. The deployment and provisioning process is simplified through features such as self-organizing networks and easy integration with OSS systems. The RBS 6401 supports both LTE and WCDMA as well as WiFi, and provides high capacity through features such as dual carriers and support for a large number of scheduled users.
Descripción de los conceptos fundamentales de redes ópticas pasivas (PON): elementos de la red de acceso óptica, topologías, variantes FTTx, protocolos (TDM y TDMA), tecnologías y estándares (BPON; GPON y EPON) y beneficios.
2. 1.BELİRSİZ İNTEGRAL
2.BELİRSİZ İNTEGRALİN ÖZELLİKLERİ
3.İNTEGRAL ALMA KURALLARI
4.İNTEGRAL ALMA METODLARI
Değişken Değiştirme (Yerine Koyma)Metodu
Kısmi İntegrasyon Yöntemi
Basit Kesire Ayırma metodu
5.TRİGONOMETRİK DÖNÜŞÜMLER YARDIMIYLA ÇÖ
6.BAZI ÖZEL DEĞİŞKEN DEĞİŞTİRMELER
7.DEĞERLENDİRME TESTİ
3. • Tanım: f : [ a , b] → R tanımlı iki fonksiyon olsun.Eğer F(x)
in türevi f(x) veya diferansiyeli f(x).dx olan F(x)
fonksiyonuna,f(x) fonksiyonunun belirsiz integrali denir ve
∫ f (x).dx = F(x) + c biçiminde gösterilir.
∫
•∫ f ( x ).dx = F( x ) + c eşitliğinde; işaretine,integral işareti,f(x) e
integrand(integral altındaki fonksiyon),f(x).dx e diferansiyel
çarpanı,F(x) e f(x) in ilkel fonksiyonu ve C ye integral
sabiti denir.
4. 1.Bir belirsiz integralin türevi,integrali alınan fonksiyona
eşittir:
∫ ( '
)
f ( x ).dx = (F( x ) + C)' = f ( x )
2.Bir belirsiz integralin diferansiyeli,integral işaretinin
(∫ f ( x).dx )
altındaki ifadeye eşittir: ′
d = f ( x ).dx
3.Bir fonksiyonun diferansiyelinin belirsiz integrali,bu
fonksiyon ile bir C sabitini toplamına eşittir:
∫ d ( f ( x)) = f ( x) + c
5. ∫4 x
5
Örnek-1- .dx belirsiz integralinin türevini bulunuz.
Çözüm :
d
dx
(∫4 x 5
)
.dx = x 5
4
Örnek-2-
∫ d ( x 3 + x) belirsiz integralini bulunuz.
∫d (x + x) = x + x + c
3 3
Çözüm :
Örnek-3- ∫ x 2 +1.dxbelirsiz integralinin diferansiyelini
bulunuz.
Çözüm :
∫ x 2 +1.dx = x 2 +1.dx
6. 1
1. ∫ x dx =
n
x n +1 +c (n ≠ −1)
n +1
1
2. ∫e .dx = e +c
x x
3. ∫ dx = ln x + c
x
4. ∫a x .dx = 1 a x +c (a > 0, a ≠ 1)
ln a
5. ∫ sin x.dx = − cos x + c 6. ∫ cos x.dx = sin x + c
∫
7. ∫ tan x. sec x.dx = sec x + c 8. cot x. cos ecx.dx = − cos ecx + c
7. 1
9. ∫ sec xdx = ∫ dx = ∫ (1 + tan 2 x)dx = tan x + c
2
cos 2 x
1
∫ cos ec xdx = ∫ 2 dx = ∫ (1 + cot 2 x)dx = − cot x + c
2
10. sin x
1
11. ∫1 + x 2 dx = arctan x +c
1
12. ∫ 1 −x 2
dx =arcsin x +c
8. Örnek-1-
∫x 5 dx belirsiz integralini bulunuz.
1 6
Çözüm: I =∫x dx = x +c
5
6
Örnek-2- ∫ (e 3 +e x ) dx belirsiz integralini bulunuz.
Çözüm: I = ∫ (e 3 + e x )dx = e 3 .x + e x + c
x5 + x 4 − 2x
Örnek-3-
∫ x5 dx belirsiz integralini bulunuz.
1 2 dx
Çözüm:
I = ∫ 1 + − 4 .dx = x + ln x − 2.∫ 4 = x + ln x − 2.∫ x − 4 dx
x x x
x −3 2
= x + ln x − 2. = x + ln x + 3 + c
−3 3x
9. 3−x
Örnek-4- ∫ 3 x 1 +
dx belirsiz integralini bulunuz.
x
x 1 3x
Çözüm: I = ∫ 3 + dx = + ln x + c
x ln 3
Örnek-5- ∫ tan 2 xdx belirsiz integralini bulunuz.
Çözüm: I =
∫ (1 + tan x − 1) dx = ∫ (1 + tan 2 x ) dx − ∫ dx = tan x − x + c
2
∫
Örnek-6- cot 4 xdx integralini hesaplayınız.
Çözüm: cot 4 xdx = cos 4 x 1 4 cos 4 x 1 (sin 4 x)1
∫ ∫ sin 4 x dx = 4 ∫ sin 4 x dx = 4 ∫ sin 4 x
1
= ln sin 4 x + c
4
10. ∫ f ( g (x ) ) g ' ( x)dx İntegralinde u=g(x) ve u ' = g ' ( x)dx
Dönüşümü yapılarak integral ∫ f ( x)du haline getirilir.
Örnek-1- ∫ ( x 4 − 2 x 2 + 3).( x 3 − x).dx integralini hesaplayınız
Çözüm: u = x − 2 x + 3
4 2
⇒ du = (4 x − 4 x).dx
3
du
du = 4( x − x).dx
3
4
= ( x 3 − x ).dx
du 1 3 1 u4 1 4
I = ∫ u 3 = ∫ u .du = +c I = ( x − 2 x 2 + 3) 4 + c
4 4 4 4 16
11. ∫e
sin x
Örnek-2- . cos x.dx integralini hesaplayınız.
Çözüm: u = sinx du = cosx.dx
I = ∫ e .du = e + c
u u
x
Örnek-3-
∫1 + x 2 dx integralini hesaplayınız.
Çözüm: u = 1+ x 2 du = 2xdx ⇒ du =x.dx
2
du
I =∫ 2 = 1 ln u + c = 1 ln(1 + x 2 ) + c
u 2 2
12. ln x
Örnek-4-
∫ x
dx integralini hesaplayınız.
Çözüm: u = ln x du =
1
x
dx
3
1
u 2
I =∫ u du = ∫ u du =
2
+c
3
2
3
2
= (ln x ) + c 2
3
13. dx
Örnek-5- ∫ e x + 1 dx integralini hesaplayınız.
Çözüm:
dx ex + 1− ex ex + 1 ex ex
I = ∫ x dx = ∫ dx = ∫ x dx − ∫ x dx = ∫ dx − ∫ x dx
e +1 e +1
x
e +1 e +1 e +1
ex
I2 = ∫ x dx u = e + 1 ⇒ du = e .dx
x x
e +1
du
I2 = ∫ = ln u + c
u
I = x − ln e +1 + c
x
14. x
e
Örnek-6- ∫ x
dx integralini hesaplayınız.
Çözüm: 1 1
u= x du = dx 2du = dx
2 x x
I = ∫ e u .2du = 2 ∫ e u du = 2e u + c I = 2e + c x
Örnek-7- ∫ sin x. cos x.dx integralini hesaplayınız.
Çözüm: u = sin x du = cos x.dx
2 2
u sin x
I = ∫ u.du = +c I= +c
2 2
15. Örnek-8- ∫ x 2 − 4 x ( x − 2).dx integralini hesaplayınız.
Çözüm: u = x2 − 4x du = (2 x − 4 x).dx = 2( x − 2).dx
du
= ( x − 2).dx
2
3
3
du 1 1u 1 7 2
I= u = ∫ u .du = +c = u +c
2 2 2 3 3
2
3
1 2
I = ( x − 4 x) 2 + c
3
16. arctan x
Örnek-9- ∫ dx integralini hesaplayınız.
1+ x 2
u = arctan x du = 1
Çözüm: dx
1+ x 2
2
u arctan 2 x
I = ∫ u.du = + c I= +c
2 2
e x + e− x
Örnek-10-
∫ e x − e− x dx integralini hesaplayınız.
−x −x
u = e −e x
du = (e + e )dx
x
du −x
I =∫
u
= ln u + c I = ln e − e + c
x
17. Örnek-11-
∫ (cot x − tan x)dx integralini hesaplayınız.
Çözüm:
∫ cot xdx − ∫ tan xdx
I1 I2
u = sin x t = cos x
du = cos x dt = − sin x.dx
I = ln u + ln t + c
I = ln sin x + ln cos x + c
18. sin 2 x
Örnek-12- ∫ 3 + cos2 x dx integralini hesaplayınız.
Çözüm: u = 3 + cos 2 x du = − 2 cos x sin x = − sin x
du
I = ∫− = − ln u + c I = − ln 3 + cos 2 x + c
u
Örnek-13- ∫ (tan 4 x + tan 2 x )dx integralini hesaplayınız.
Çözüm: I = ∫ tan x(tan x + 1)dx
2 2
u = tan x du = (1 + tan 2 x)dx
u 3
tan 3 x
I = ∫ u du = + c
2
I= +c
3 3
19. dx
Örnek-14- ∫ 9 − 25 x 2
integralini hesaplayınız.
Çözüm:
dx 1 bx
a, b ∈ R − { 0} ⇒ ∫ a −b x
2 2 2
= arcsin + c
b a
dx 1 5x
∫ 9 − 25 x 2
= arcsin + c
5 3
20. ∫u.dv =u.v −∫v.du
u ve v ' yi seçerken;
1. dv’nin integralinden v kolayca bulunabilir.
2. ∫ du integralini hesaplamak ∫u.du
v.
integralinden daha kolay olmalı.
2. u seçimi yaparken öncelik sırası :
L A P T Ü
logoritma arc polinom trigonometrik üstel
21. ∫ x.e .dx
x
Örnek-1- integralini hesaplayınız.
Çözüm:
u=x dv = e .dx x
du = dx v=e x
∫ x.e .dx = x.e − ∫ e .dx
x x x
= x.e −e + c
x x
22. Örnek-2-
∫ ln x.dx integralini hesaplayınız.
Çözüm:
u = ln x dv = dx
1
du = dx v=x
x
1
∫ lnx.dx = x.lnx - ∫ x. x dx
= x.lnx - x + c
23. Örnek-3- ∫ e x . sin x.dx integralini hesaplayınız.
Çözüm:
u = sin x dv = e x dx
du = cos x.dx v = ex
sin x.e x − ∫ e x . cos x.dx
u = cos x dv = e x .dx
du = − sin x.dx v=e x
(
I = e x . sin x − e x . cos x − (e x cos x − ∫ − e x . sin x.dx )
I = e x . sin x − e x . cos x − ∫ e x . sin x.dx
I
ex
2 I = e .( sin x − cos x ) = .( sin x − cos x ) + c
x
2
24. ∫ (
Örnek-4- ln x + )
x 2 −1 .dx integralini hesaplayınız.
Çözüm:
(
u = ln x + x −1 2
) dv = dx
1
du = .dx v =x
x −1
2
(
I = x.lnx x + x −1 − ∫ 2
) x.dx
x 2 −1
(
I = x. ln x + x −1 − x −1 + c
2
) 2
25. Örnek-5- ∫ cos( ln x ).dx integralini hesaplayınız.
Çözüm:
u = cos( ln x ) dv = dx
1
du = -sinlnx. dx v=x
x
lnx
I = cos(lnx).x - ∫ - sin xdx
x
u = sin(lnx) dv = dx
1
du = cos(ln x).dx v = x
x
I = x.cos(lnx) + x.sin(lnx) - ∫ cos(lnx)
x I
I = ( cos(ln x) + sin(ln x) )
2
26. P ( x).dx
∫ Q( x) integralinde der[ p(x)] < der[Q( x)] ise Q( x)
çarpanlarına ayrilir .
der ( P ( x) ) ≥ der ( Q( x) ) ise adi bölme ile
P(x) K ( x)
= B( x) + haline getirilir.
Q(x) Q( x)
27. x3 + 2 x 2 + x + 2
Örnek-1- ∫ x+1
dx integralini hesaplayınız.
Çözüm:
x3 + 2 x2 + x + 2
3 2 X+1
x +x x2 + x
x2 + x + 2 2
- x +x+
2
- x +x
2
x +1
2
3 2
2 2 x x 2dx
I = ∫x + x+ .dx = + + ∫
x +1 3 2 x +1
3 2
x x
I = + + 2 ln x + 1 + c
3 2
28. x −1
Örnek-2- ∫ x.( x + 1)dx integralini hesaplayınız.
Çözüm: x −1 A B
=− + x - 1 = A(x + 1) + B(x)
x.( x + 1) x x+1
x -1 1 2
x = 0 için A = -1 =− +
x.(x + 1) x x+1
x = -1 için B = 2
1 2
∫ − x + x + 1 dx = − ln x + 2 ln x + 1 + c
( x + 1) 2
I = ln +c
x
29. dx
Örnek-3- ∫ x.(x − 1) 2 integralini hesaplayınız.
Çözüm: 1 A B C
= + +
x.( x −1) 2
x x −1 ( x −1) 2
1 = A( x −1) + Bx( x −1) + Cx
2
x =1 için C =1, x = o için A =1
x = 2 için B = -1
dx 1 −1 1
∫ x.(x - 1) 2 = ∫ ( x + x −1 + ( x −1) 2 )dx
1
I = ln x − ln x −1 − +c
x −1
30. dx
Örnek-4-
∫ x 2 −16 integralini hesaplayınız.
Çözüm: dx dx
∫ x 2 − 16 = ∫ ( x − 4).( x + 4)
1 A B
= +
( x − 4).( x + 4) x − 4 x + 4
1 = A( x + 4) + B ( x − 4)
1
x = −4 için B = −
8
1
x =4 için A =
8
1 1
−
dx 1 x −4
8 + 8 dx = ln
∫ x 2 −16 = ∫ x − 4 x + 4 8 x + 4 + c
31. sin 2 x = 2 sin x. cos x
1. sin 2 x + cos 2 x = 1 3.
2. sec 2 x − tan 2 x = 1 4. cos 2 x = 2. cos 2 x − 1
= 1− 2 sin 2 x
1
*
sin a. sin b = − [ cos(a + b) − cos(a − b)]
2
1
* sin a. cos b = [ sin( a + b) + sin( a − b)]
2
1
* cos a. cos b = [ cos(a + b) + cos(a − b)]
2
32. ∫ sin ax. sin bx, ∫ sin bx. cos bx, ∫ cos ax. cos bx
BİÇİMİNDEKİ İNTEGRALLER
Örnek: ∫ cos 4 x.cos 2 x.dx integralini hesaplayınız.
Çözüm:
1 11 1
I = ∫ (cos 6 x + cos 2 x).dx = sin 6 x + sin 2 x + c
2 26 2
1 1
I = sin 6 x + sin 2 x + c
12 4
33. ∫ sin x.dx, ∫ cos x.dx
n n
BİÇİMİNDEKİ İNTEGRALLER
Örnek-1- ∫ sin 2 x.dx integralini hesaplayınız.
Çözüm:
1 − cos 2 x 1 1
∫ sin x.dx = ∫ 2 dx = ∫ 2dx − 2 ∫ cos 2 x.dx
2
1 1
= sin 6 x + sin 2 x + c
12 4
34. ∫ sin
4
Örnek-2- x.dx integralini hesaplayınız.
Çözüm:
2
1 − cos 2 x 1
∫ sin x.dx = ∫ (sin x) .dx = ∫ dx = ∫ (1 − cos 2 x)2 .dx
4 2 2
2 4
1 1 1
= ∫ (1 − 2 cos 2 x + cos 2 x)dx = ( x − 2. sin 2 x + ∫ cos 2 xdx)
2 2
4 4 2
1 + cos 4 x
=
2
x 1 1 1 + cos 4 x x 1 1 1
I = − sin 2 x + ∫ dx = − sin 2 x + ( x + sin 4 x) + c
4 4 4 2 4 4 8 4
3 x sin 2 x 1
I= − + sin 4 x + c
8 4 32
35. 5
Örnek-3- sin xdx integralini hesaplayınız.
Çözüm:
sin xdx = ∫ (sin x) .sin x.dx = ∫ (1 − cos x) .sin x.dx
5 2 2 2 2
du = − sin xdx
u = cos x I = ∫ (1 − u ) .(− du )
2 2
− du = sin x.dx
I = ∫ (1 − 2u + u ).(−du ) = ∫ (−1 + 2u − u ).du
2 4 2 4
2 3 u5 2 3 1 5
I = −u + u − I = − cos x + cos x − cos x + c
3 5 3 5
36. ∫ sin n x. cos m x.dx
BİÇİMİNDEKİ İNTEGRALLER
Örnek:-1-
∫ sin 2 x. cos3 x.dx integralini hesaplayınız.
Çözüm:
∫ sin x. cos x.dx = ∫ sin x. cos x. cos x.dx
2 3 2 2
= ∫ sin x.(1 − sin x). cos x.dx
2 2
u = sin x du = cos x.dx
I = ∫ u 2 .(1 − u 2 ).du = ∫ (u 2 − u 4 ).du
3 5 3 5
u u sin x sin x
= − +c= − +c
3 5 3 5
37. x. sin 3 x.dx ∫ cos
Örnek-2- 4 integralini hesaplayınız.
Çözüm: 4
x. sin 3 x.dx = ∫ cos 4 x. sin 2 x. sin x.dx ∫ cos
= ∫ cos 4 x .(1 − cos 2 x). sin x.dx u = cos x du = − sin x.dx
I = ∫ u .(1 − u ).(− du ) = ∫ (−u + u ).du
4 2 4 6
− u5 u7 − cos 5 x cos 7 x
I= + +c = + +c
5 7 5 7
38. ∫ tan x.dx, ∫ cot x.dx
n n
BİÇİMİNDEKİ İNTEGRALLER
Örnek-1- ∫ tan x.dx integralini hesaplayınız.
Çözüm:
sin x
∫ tan x.dx = ∫ cos x .dx
u = cos x du = −sin x.dx
I = − ln cos x + c
39. ∫ tan x.dx
Örnek-2- 2 integralini hesaplayınız.
Çözüm:
∫ tan x.dx = ∫ (tan x + 1 − 1).dx
2 2
= ∫ (tan x + 1).dx − ∫ dx
2
I = tan x − x + c
40. Örnek-3-
∫ tan 4 x.dx integralini hesaplayınız.
Çözüm:
∫ tan 4 x.dx = ∫ tan 2 x. tan 2 x.dx = ∫ (sec 2 − 1). tan 2 x.dx
∫ sec 2 . tan 2 x.dx − ∫ tan 2 x.dx
u = tan x 2
= ∫ (tan x + 1 − 1)dx
du = sec 2 x
I = ∫ u 2 .du − ∫ tan 2 ( x + 1).dx + ∫ dx
3 3
u tan x
= − tan x + x + c = − tan x + x + c
3 3
41. ∫ tan x.dx
5
Örnek-4- integralini hesaplayınız.
Çözüm:
I = ∫ tan x.dx = ∫ tan x. tan x.dx = ∫ tan x.(sec x − 1).dx
5 3 2 3 2
= ∫ tan x. sec x.dx − ∫ tan x.dx
3 2 3
u = tan x du = sec 2 x.dx
tan 2 x tan 4 x tan 2 x
I = ∫ u .du −
3
− ln cos x + c = − − ln cos x + c
2 4 2
42. ∫ sec x. tan
n m
x.dx
BİÇİMİNDEKİ İNTEGRALLER
Örnek-1- ∫ sec x. tan 3 x.dx integralini hesaplayınız.
Çözüm:
∫ sec x. tan x.dx = ∫ tan x. sec x. tan x.dx
3 2
= ∫ (sec x −1). sec x. tan x.dx
2
u = secx du = secx.tanx.dx
u3 sec 3 x
I = ∫ (u 2 −1)du = −u + c = − sec x + c
3 3
43. ∫ sec
6 3
Örnek-2- x. tan x.dx integralini hesaplayınız.
Çözüm:
∫ sec 6 x. tan 3 x.dx = ∫ sec5 x. tan 2 x. sec x. tan x.dx
u = secx du = secx.tanx.dx
I = ∫ u 5 (u 2 − 1).du = ∫ (u 7 − u 5 ).du = ∫ u 7 .du − ∫ u 5 .du
sec8 x sec 6 x
I= − +c
8 6
44. İNTEGRALİNDE sinx VE cosx' in RASYONEL OLARAK
BULUNDUGU İNTEGRALLER
x
u = tan
2
1+ u2
4 2u
x sinx =
2
1+ u 2
2
1- u
1 cosx =
1+ u 2
2du
dx =
1+ u 2
45. dx
Örnek-1-
∫ 1 + sin x − cos x integralini hesaplayınız.
Çözüm: 2du
dx 1+ u2 du
∫ 1 + sin x − cos x = 2u 1 − u 2 = u (u + 1)
1+ −
1+ u 1+ u2
2
A B
I = + 1=A(u + +
1) B(u)
u u +1
u(A +B) ⇒+ =
A B 0
A =1 B =-1
du du
I =∫u +−∫ u += u − u +
1
ln ln 1
x
tan
u 2
=ln + =
c ln +c
u +1 x
tan + 1
2
46. dx
Örnek-2-
∫ 2 − sin x integralini hesaplayınız.
Çözüm:
2du
dx 1 +u 2 du
∫ 2 −sin x = ∫ 2 + 2u 2 − 2u = ∫ u 2 −u +1
1 +u 2
du 1 du
= 2
− +1 = ∫ 2
1 u 1 3
2
u − u − +
⇒
2 2 2
47. du 1 u
∫ u 2 + a 2 = a arctan a
x
2 2 tan − 1
I= arctan 2 +c
3 3
48. a −b x
2 2 2 ‘den başka köklü ifade bulundurmayan
integralleri hesaplamak için
a ∏ ∏
x = sin u - ≤ u ≤
b 2 2
Değişken değiştirmesi yapılır.
49. Örnek:
bulunuz.
∫ 4 − 9 x 2 .dx integralinin değerini
2 2
Çözüm: x = sin u deg.deg. dx = cosu.du
3 3
2
2
4 - 9x = 4 - 9 sin u = 4 - 4sin 2u = 2 cos u
2
3
2 4 1 + cos 2u
I = ∫ 2 cos u. cos u.du = ∫ .du
3 3 2u
2 2 1
I = ∫ (1 + cos 2u ) du = u + sin 2u + c
3 3 2
2 3x 1 3x
I = arcsin + 4 − 9x + c
2
3 2 3 2
50. a 2 + b 2 x 2 ‘den başka köklü ifade bulunmayan
integralleri hesaplamak için
a ∏ ∏
x = tan u - ≤ u ≤
b 2 2
Değişken değiştirmesi yapılır.
51. dx
Örnek: ∫ 4 + x2
integralini hesaplayınız.
Çözüm:
x = 2 tan u deg.deg. yap. dx = 2(1 + tan u ) = 2. sec du
2 u
4 + x = 4 + 4 tan u = 2 1 + tan u = 2 sec u = 2 sec u
2 2 2 2
sec x(sec x + tan x) (sec 2 x + sec x. tan x)dx
∫ sec x.dx = ∫ sec x + tan x dx = ∫ sec x + tan x
u = secx + tanx
du = (secx - tanx + sec 2 x)dx
du
I = ∫ = ln u + c = ln sec x + tan x + c
u
52. b x −a
2 2 2
‘den başka köklü ifade bulundurmayan
integralleri hesaplamak için:
a
x = sec u
b
Değişken değiştirmesi yapılır.
53. 9 x 2 −1
Örnek:
∫ x
dx integralini hesaplayınız.
Çözüm: 1 1
x = sec u dx = sec u. tan u.du
3 3
1
9 x −1 =9.
2
sec 2 u −1 = tan u
9
1
tan u sec u. tan u.du
I =∫ 3 = ∫ tan 2 u.du
1
sec u
3
I = tan u −u +c = 9 x 2 −1 −u +c
I = 9 x 2 −1 −arctan ( 9 x 2 −1 +c)
54. cos x − sin x
1. ∫ sin x + cos x dx belirsiz integrali için
Aşağıdakilerden hangisi doğrudur?
A) I = sin x + cos x + c
B) I = 2. sin x + c
C) I = 2 sin x + cos x + c
D) I = sin x. cos x + c
E) I = 2 sin x. cos x + c
55. x 4 +4
2. ∫ x 4 dx Belirsiz integrali aşağıdakilerden
hangisi olamaz?
3x 4 + x 3 −
3 4
A) 3x 3
3x 4 + x 3 −
4 4
B) 3x 3
3x 4 + x 3 − x
5 4
C 3x 3
3x 4 + x 3 −
6 4
D) 3x 3
3x 4 −4
E)
3x 3
56. 3.
∫ sin 2 x. cos 2 x.dx İntegralinin çözümü aşağıdakilerden
hangisidir?
x sin 4 x
A) I= + +c
8 32
3x sin 4 x
B) I = − +c
8 32
x sin 4 x
C) I =− − +c
8 32
D) I = x −sin 4 x +c
8 32
E) I = −3x −sin 4 x +c
8 32
57. 2 x3 + 2 x + 1
4 ∫ dx Belirsiz integrali için aşağıdakilerden
x +1
2
hangisi doğrudur?
x 2 + arctan x + c
A)
x 3 + arctan x + c
B)
C) x 2 + ln( x 2 + 1) + c
1
D) x + ln( x 2 + 1) + c
2
E)
x + arctan( x + 1) + c
2 2
58. ln x
5. ∫ x 2 dx belirsiz integrali için aşağıdakilerden
hangisi doğrudur?
ln x 1
I= − +c
A) x x
ln x 1
I= + +c
B) x x
−ln x 1
I= + +c
C) x x
1
D) I =ln − +c
x
ln x 1
E) I =− − +c
x x
59. 6. ∫ 8 sin x. cos x. cos 2 x.dx belirsiz integrali için
Aşağıdakilerden hangisi doğrudur?
A) cos 4 x +c
−cos 4 x +c
B)
1
sin 4 x +c
C) 4
cos 8 x
D) − 2 +c
E) − cos 4 x +c
2
60. ∫ sin x.dx
7. 3 integralinin değeri aşağıdakilerden
hangisidir?
cos 3 x
I =−cos x + +c
A) 3
cos 3 x
B) I =cos x + +c
3
sin 3 x
C) I =sin x + +c
3
cos 3 x
D) I =sin x + +c
3
E) cos 3 x
I =−sin x + +c
3
61. dx
8. ∫ x2 + x belirsiz integrali için, aşağıdakilerden
hangisi doğrudur?
1
ln +c
A) x+ 1
1
ln +c
B) x +
2
1
ln x 2 + +
x c
C)
x
ln +c
D) x+ 1
x2
E) ln +c
x+ 1
62. 9. ∫ − sin(sin 2 x). sin 2 x.dx aşağıdakilerden hangisidir?
A) cos(sin x) + c
B) sin(sin 2 x) + c
C) cos(sin x) + c
D) cos(sin 2 x) + c
E) cos x + c
63. 3x + x − 2
2
10.∫ dx integralinin değeri
( x −1)( x +1)
2
aşağıdakilerden hangisidir?
A) I = − ln x − 1 + ln x 2 + 1 + 3 arctan x + c
B) I = ln x − 1 + ln x 2 + 1 + 3 arctan x + c
C) I = ln x − 1 − ln x 2 + 1 + 3 arctan x + c
D) I = ln x − 1 − ln x 2 + 1 − 3 arctan x + c
E) I = − ln x − 1 + ln x 2 + 1 − 3 arctan x + c