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General Discussion => Other Discussions => Math and Science => Topic started by: Sorunome on January 24, 2013, 08:20:55 pm

Title: Trig Substitution
Post by: Sorunome on January 24, 2013, 08:20:55 pm

Hey, today in ap calc we learned trig substitution, and i understand everything up to the part where you have to substitute the theta back into so that your awnser is in terms of x. Could somebody please try to explain it to me?

Title: Re: Trig Substitution
Post by: AngelFish on January 24, 2013, 10:02:30 pm

The reason you substitute theta back is is because the original substitution mapped the function to a polar domain. The answer is correct for THAT domain, but it's not necessarily correct for the Euclidean domain the original problem was phrased in. So, what you do is remap the function back to a euclidean domain by substituting theta.

Is that what you were asking?

Title: Re: Trig Substitution
Post by: Sorunome on January 24, 2013, 10:04:00 pm

oh, no, i know WHY you need to substitute it back but i don't get HOW you do it.

Title: Re: Trig Substitution
Post by: ruler501 on January 24, 2013, 10:21:38 pm

You use SineOppositeHypotenuse CosineAdjacentHypotenuse TangentOppositeAdjacent
If you subsituted sine in you need to solve for it in terms of x. Then you phrase it as a fraction (if its not a fraction just put a 1 under it). Then match sides using the trig functions. (Opposite top, Hypotenuse bottom for sine) draw a triangle and solve it in terms of x. You can then substitute that into any trig function to solve for it in terms of x. If you want an example I can work one out later.

Title: Re: Trig Substitution
Post by: Sorunome on January 24, 2013, 10:32:55 pm

Erm, i don't get what you mean.
We had e.g. for
integral(1/sqrt(16-x^2)) then in the end
theta+c

and how to get from that to
arcsin(x/4)+c
+c is obvious :P

Title: Re: Trig Substitution
Post by: Yeong on January 24, 2013, 11:03:22 pm

Memorizing formula, of course! :P

integral(1/sqrt(a^2-x^2))=arcsin(x/a)

Title: Re: Trig Substitution
Post by: Sorunome on January 24, 2013, 11:03:42 pm

i mean without dat :P
so the hard way :P

Title: Re: Trig Substitution
Post by: Yeong on January 24, 2013, 11:12:25 pm

integral(1/sqrt(16-x^2))dx

(http://img.removedfromgame.com/imgs/triangle.png)

using trig substitution

sin(-)=x/4
4sin(-)=x

sqrt(4^2-x^2)=4cos(-)

dx = d/dx(x)
dx = d/dx(4sin(-))
dx = 4cos(-)d(-)

substitute to original equation:

integral((1/4cos(-))(4cos(-)d(-)))
= integral (1 d(-))

integrate it and you get (-)+C.

now since sin(-)=x/4
(-) = arcsin(x/4)

therefore (-)+C = arcsin(x/4)+C

Title: Re: Trig Substitution
Post by: Sorunome on January 25, 2013, 06:52:01 pm

Ok, and what did you need the triangle drawing for now?

Title: Re: Trig Substitution
Post by: aeTIos on January 26, 2013, 07:42:12 am

He used the triangle drawing to not have to explain what kind of triangle he was using. (as in, side length and stuff)

Title: Re: Trig Substitution
Post by: Sorunome on January 26, 2013, 03:56:32 pm

i mean, sqrt(16-x²) is nohere used

Title: Re: Trig Substitution
Post by: lkj on January 26, 2013, 04:01:55 pm

Quote

sqrt(4^2-x^2)=4cos(-)

:P

Title: Re: Trig Substitution
Post by: Sorunome on January 26, 2013, 04:03:59 pm

Quote from: lkj on January 26, 2013, 04:01:55 pm

Quote

sqrt(4^2-x^2)=4cos(-)

:P

oh, i didn't see that one......mhmm, maybe i get it now, idk......