How to integrate int F sin^2(At) dt

  • Thread starter danago
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  • #1
danago
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Hey. Ive barely started learning about integration, but during an experiment we conducted for physics, the following came up:

[tex]
\int {F\sin ^2 At} dt
[/tex]

How would i integrate that? I do understand that it would generally require that i show my working on the issue, but in this case, its a bit hard, since i dont have any working.

If someone could briefly explain how i would go about evaluating this indefinite integral, i would be greatly appreciative.

Thanks,
Dan.
 

Answers and Replies

  • #2
TD
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I assume F and A are constants, so that shouldn't be of any trouble.
The annoying part is the square of the sine, but you could use:

[tex]
\cos \left( {2x} \right) = \cos ^2 x - \sin ^2 x = 1 - 2\sin ^2 x \Rightarrow \sin ^2 x = \frac{{1 - \cos \left( {2x} \right)}}{2}
[/tex]

Does that help?
 
  • #3
danago
Gold Member
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Hmmm...ok i understand that. How would i integrate something like this:

[tex]
\int {\cos At} {\rm dt}
[/tex]

Would it just be [tex]-sin At+c[/tex]
?
 
  • #4
TD
Homework Helper
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Almost, you have to watch out with the A. Hint: dt = 1/A d(At).
Or, use an explicit substitution: let y = At so dy = Adt so 1/A dy = dt.
 
  • #5
J77
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...and you may want to rethink that minus sign.
 
  • #6
danago
Gold Member
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ahh so would this be right? :

[tex]
\int {\cos At} {\rm dt = }\frac{{\sin At}}{A} + c
[/tex]
 
  • #7
J77
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Yep - check by differentiating.
 
  • #8
danago
Gold Member
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Awesome :D Thanks

OK so back to the original question. Is this working right?:

[tex]
\displaylines{
\int {F\sin ^2 At} {\rm }dt = F\int {\sin ^2 At} {\rm }dt \cr
= F\int {\frac{{1 - \cos (2At)}}{2}} {\rm }dt \cr
= \frac{F}{2}\int {1 - \cos (2At)} {\rm }dt \cr
= \frac{F}{2}\left( {\int {1{\rm }dt - \int {\cos (2At)} {\rm }dt} } \right) \cr
= \frac{F}{2}\left( {t - \frac{{\sin 2At}}{{2A}}} \right) \cr
= \frac{{Ft}}{2} - \frac{{F\sin 2At}}{{4A}} \cr}
[/tex]
 
  • #9
J77
1,083
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yep - plus constant, of course.
 
  • #10
danago
Gold Member
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Oops forgot about that.

Anyway, thanks very much for the help. Greatly appreciated :D
 

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