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|<math> \int \frac {1}{\operatorname{ch}\ u}  \ d u = \arcsin{\left ( \operatorname{th}\,u \right )} \qquad 2 arc th  e^u</math>
 
|<math> \int \frac {1}{\operatorname{ch}\ u}  \ d u = \arcsin{\left ( \operatorname{th}\,u \right )} \qquad 2 arc th  e^u</math>
 
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|<math> \int \frac {1}{\operatorname{sh}\ u}  \ d u = \ln \operatorname{th}\,\frac{2}{2} \qquad Arg \coth e^u</math>
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|<math> \int \frac {1}{\operatorname{sh}\ u}  \ d u = \ln \operatorname{th}\,\frac{2}{2} \qquad - Arg \coth e^u</math>
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|<math> \int \frac {1}{\operatorname{ch^2}\ u}  \ d u =  \operatorname{th}\,u  </math>
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|<math> \int \frac {1}{\operatorname{sh^2}\ u}  \ d u = - \operatorname{coth}\,u  </math>
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|<math> \int \operatorname{th^2}\ u \ d u = u - \operatorname{th}\,u  </math>
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|<math> \int \operatorname{coth^2}\ u \ d u = u - \operatorname{coth}\,u  </math>
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|<math> \int \operatorname{sh^2}\ u \ d u = \frac {\operatorname{sh}\,{2 u}}{4} - \frac{u}{2}=\frac{1}{2}\left ( \operatorname{sh}\,u \ \operatorname{ch}\,u  - u \right )</math>
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|<math> \int \operatorname{ch^2}\ u \ d u = \frac {\operatorname{sh}\,{2 u}}{4} + \frac{u}{2}=\frac{1}{2}\left ( \operatorname{sh}\,u \ \operatorname{ch}\,u  + u \right )</math>
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|<math> \int \frac{\operatorname th \ u}{\operatorname ch \ u} \ d u = - \frac {1}{\operatorname ch \, u }</math>
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|<math> \int \frac{\operatorname coth \ u}{\operatorname sh \ u} \ d u = - \frac {1}{\operatorname sh \, u }</math>
 
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! style="background-color: rgb(238, 238, 238); background-image: none; background-repeat: repeat; background-attachment: scroll; background-position: 0% 0%; -moz-background-size: auto auto; -moz-background-clip: -moz-initial; -moz-background-origin: -moz-initial; -moz-background-inline-policy: -moz-initial; font-size: 110%;" colspan="2" | Important Transformations
 
! style="background-color: rgb(238, 238, 238); background-image: none; background-repeat: repeat; background-attachment: scroll; background-position: 0% 0%; -moz-background-size: auto auto; -moz-background-clip: -moz-initial; -moz-background-origin: -moz-initial; -moz-background-inline-policy: -moz-initial; font-size: 110%;" colspan="2" | Important Transformations

Revision as of 08:07, 19 November 2010

Table of Infinite Integrals
General Rules
$ \int a d x = a x $
$ \int a f ( x ) d x = a \int f ( x ) d x $
$ \int ( u \pm v \pm w \pm \cdot \cdot \cdot ) d x = \int u d x \pm \int v d x \pm \int w d x \pm \cdot \cdot \cdot $
$ \int u d v = u v - \int v d u $
$ \int f ( a x ) d x = \frac{1}{a} \int f ( u ) d u $
$ \int F { f ( x ) } d x = \int F ( u ) \frac{dx}{du} d u = \int \frac{F ( u )}{f^' ( x )} d u \qquad u = f ( x ) $
$ \int u^n d u = \frac{u^{n+1}}{n+1} \qquad n \neq -1 $
$ \int \frac{d u}{u} = \ln u \ ( if \ u > 0 ) \ or \ln {-u} \ ( if \ u < 0 ) = \ln \left | u \right | $
$ \int e^u d u = e^u $
$ \int a^u d u = \int e^{u \ln a} d u = \frac{e^{u \ln a}}{\ln a} = \frac{a^u}{\ln a} \qquad a > 0 \ and \ a \neq 1 $
$ \int \sin u \ d u = - \cos u $
$ \int \cos u \ d u = \sin u $
$ \int \tan u \ d u = - \ln {\cos u} $
$ \int \cot u \ d u = \ln {\sin u} $
$ \int \frac{d u}{\cos u} = \ln { \left ( \frac{1}{\cos u} + \tan u \right )} = \ln{\tan {\left ( \frac{u}{2}+\frac{\pi}{4}\right )}} $
$ \int \frac{d u}{\sin u} = \ln { \left ( \frac{1}{\sin u} - \cot u \right )} = \ln{\tan { \frac{u}{2}}} $
$ \int \frac{d u}{\cos ^2 u} = \tan u $
$ \int \frac{d u}{\sin ^2 u} = - \cot u $
$ \int \tan ^2 u \ d u = \tan u - u $
$ \int \cot ^2 u \ d u = - \cot u - u $
$ \int \sin ^2 u \ d u= \frac{u}{2} - \frac{\sin {2 u}}{4} = \frac{1}{2}\left( u - \sin u \cos u \right ) $
$ \int \frac {1}{\cos u} \tan u \ d u = \frac{1}{\cos u} $
$ \int \frac {1}{\sin u} \cot u \ d u = - \frac{1}{\sin u} $
$ \int \operatorname{sh}\,u \ d u = \operatorname{ch}\,u $
$ \int \operatorname{ch}\,u \ d u = \operatorname{sh}\,u $
$ \int \operatorname{th}\,u \ d u = \ln \operatorname{ch}\,u $
$ \int \operatorname{coth}\,u \ d u = \ln \operatorname{sh}\,u $
$ \int \frac {1}{\operatorname{ch}\ u} \ d u = \arcsin{\left ( \operatorname{th}\,u \right )} \qquad 2 arc th e^u $
$ \int \frac {1}{\operatorname{sh}\ u} \ d u = \ln \operatorname{th}\,\frac{2}{2} \qquad - Arg \coth e^u $
$ \int \frac {1}{\operatorname{ch^2}\ u} \ d u = \operatorname{th}\,u $
$ \int \frac {1}{\operatorname{sh^2}\ u} \ d u = - \operatorname{coth}\,u $
$ \int \operatorname{th^2}\ u \ d u = u - \operatorname{th}\,u $
$ \int \operatorname{coth^2}\ u \ d u = u - \operatorname{coth}\,u $
$ \int \operatorname{sh^2}\ u \ d u = \frac {\operatorname{sh}\,{2 u}}{4} - \frac{u}{2}=\frac{1}{2}\left ( \operatorname{sh}\,u \ \operatorname{ch}\,u - u \right ) $
$ \int \operatorname{ch^2}\ u \ d u = \frac {\operatorname{sh}\,{2 u}}{4} + \frac{u}{2}=\frac{1}{2}\left ( \operatorname{sh}\,u \ \operatorname{ch}\,u + u \right ) $
$ \int \frac{\operatorname th \ u}{\operatorname ch \ u} \ d u = - \frac {1}{\operatorname ch \, u } $
$ \int \frac{\operatorname coth \ u}{\operatorname sh \ u} \ d u = - \frac {1}{\operatorname sh \, u } $
Important Transformations
$ \int F( a x + b) d x =\frac{1}{a} \int F( u) d x \qquad u = a x + b $

Alumni Liaison

has a message for current ECE438 students.

Sean Hu, ECE PhD 2009