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

[danago]

Hey. I've barely started learning about integration, but during an experiment we conducted for physics, the following came up:

<br /> \int {F\sin ^2 At} dt<br />

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 don't have any working.

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

Thanks,
Dan.

 

[TD]

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:

<br /> \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}<br />

Does that help?

 

[danago]

Hmmm...ok i understand that. How would i integrate something like this:

<br /> \int {\cos At} {\rm dt}<br />

Would it just be -sin At+c
?

 

[TD]

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.

 

[J77]

...and you may want to rethink that minus sign.

 

[danago]

ahh so would this be right? :

<br /> \int {\cos At} {\rm dt = }\frac{{\sin At}}{A} + c<br />

 

[J77]

Yep - check by differentiating.

 

[danago]

Awesome :D Thanks

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

<br /> \displaylines{<br /> \int {F\sin ^2 At} {\rm }dt = F\int {\sin ^2 At} {\rm }dt \cr <br /> = F\int {\frac{{1 - \cos (2At)}}{2}} {\rm }dt \cr <br /> = \frac{F}{2}\int {1 - \cos (2At)} {\rm }dt \cr <br /> = \frac{F}{2}\left( {\int {1{\rm }dt - \int {\cos (2At)} {\rm }dt} } \right) \cr <br /> = \frac{F}{2}\left( {t - \frac{{\sin 2At}}{{2A}}} \right) \cr <br /> = \frac{{Ft}}{2} - \frac{{F\sin 2At}}{{4A}} \cr}<br />

 

[J77]

yep - plus constant, of course.

 

[danago]

Oops forgot about that.

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