Common Integrals: "The Standard Integral" Explained

[UrbanXrisis]

my book integrates this using "the standard integral"

\int e^{at} cos \omega t dt = \frac{1}{a^2+ \omega^2} e^{at} (a cos \omega t+ \omega sin \omega t) +c
where a is a constant

what is the standard integral?

 

[hotvette]

I'm guessing a bit, but I believe the term "standard integral" just means common, known, ones. For example, following would be considered a standard integral:

\int x^n dx = \frac{x^{n+1}}{n+1} +c

 

[Tx]

I would say the 'standard integral' that your text is referring to would be Integration By Parts. Note: I am assuming 'w' is constant.

If you have never seen the formula here it is:
http://mathworld.wolfram.com/IntegrationbyParts.html

 

[UrbanXrisis]

ohh... wow, I feel dumb. okay, I thought it was going to be some weird trig arctan integral that I have never seen. by parts would do it. thank you!

 

[Benny]

I always thought that standard integrals are more general results which can be utilised by plugging in values specific to your problem. In this case, your integral looks like a fairly general one to me.

Anyway an alternative to integration by parts is the following:

<br /> \int {e^{\left( {a + \omega i} \right)t} } dt<br />

= \frac{1}{{a + \omega i}}e^{\left( {a + \omega i} \right)t}

<br /> = \frac{{a - \omega i}}{{a^2 + \omega ^2 }}e^{\left( {a + \omega i} \right)t} <br />

<br /> = \frac{{a - \omega i}}{{a^2 + \omega ^2 }}e^{at} \left( {\cos \left( {\omega t} \right) + i\sin \left( {\omega t} \right)} \right)<br />

<br /> \int {e^{at} \cos \left( {\omega t} \right)} dt = {\mathop{\rm Re}\nolimits} \left\{ I \right\}<br />

<br /> = \frac{1}{{a^2 + \omega ^2 }}e^{at} \left( {a\cos \left( {\omega t} \right) + \omega \sin \left( {\omega t} \right)} \right)<br />

I left out the constant of integration.