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Integral Trig Substitutions.

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Where do the equations that are given to use the certain trig function in certain cases
if you have a proof , that's what im looking for
bascally you can test them after but i'm wondering how'd you do it with a triangle before guessing

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I'm not sure what you mean, but see if the picture below helps you.
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not quite i understand this it's where they get these rules
in other words how'd they know to use sec,sin,and tan...
It depends on the question, if its \[a^2-x^2\] Then from triangle |dw:1361605801008:dw| You know that \[\sin(\theta)=\frac{x}{a} \\ \\ x=asin(\theta)\]
whered you get the triangle
basically i'm asking how do they get the triangle, is it simply guess and check or is there a way that if i don't remember these rules that i can use trig to find these
Its from \[a^2-x^2\] Another example, if its \[a^2+x^2\] Notice the plus is given for adjacent side and opposite side, Pythagoras theorem. |dw:1361606125590:dw| then you get \[x=atan(\theta)\]
I mean why do derivatives of certain cartesian functions yield a trig function
You need it only if you have something like this, you can use trig substitution for 1/sqrt(a^2+x^2). But that depends on the overall equation.
where does this come into play... for example you have the inverse of tan(x) yields a derivative of 1/x^2+1
cause simply you're telling me to remember certain terms, which seems as though i'm not thoroughly learning the subject itself. If i forget the equations, then i'm out of luck. However if i forgot them i could simply start from scratch
i'm looking for proofs =/
Ok so the derivative of the Inverse Trig fuctions can be found several ways. I like to do it using the triangle method. \[\large y=\arctan x \qquad \rightarrow \qquad x=\tan y\] If we want to, we can write this as,\[\large \tan y=\frac{x}{1}=\frac{opposite}{adjacent}\] Let's setup a triangle to represent this relationship.|dw:1361608691656:dw|
We'll take the derivative from this point, \[\large x=\tan y\]With respect to x,\[\large 1=(\sec^2 y) y'\]Solving for y',\[\large y'=\frac{1}{\sec^2y} \qquad \rightarrow \qquad y'=\cos^2y\] \[\large y'=\left(\frac{adjacent}{hypotenuse}\right)^2\] \[\large y'=\left(\frac{1}{\sqrt{1+x^2}}\right)^2\] \[\large y'=\frac{1}{1+x^2}\]
alright this makes sense as this is basically what the other guy did only using something he called the inverse something theorem which i've never heard of but it makes sense. so this is the proof for trig functions now does this come into play when you try doing trig substitution?
No not really. Even when dealing with a very straight forward integral like this one,\[\large \int\limits\frac{1}{1+x^2}dx\]I'm not usually thinking in my head, "Oh! That's the derivative of arctan!" I'm generally looking at the relationship the x^2 has with the other number on the bottom. I know that \(1+\tan^2 \theta=\sec^2\theta\), So if I can make a substitution changing the x to tangent ~ I can reduce the denominator to \(\sec^2\theta\). That will eliminate the addition in the bottom, making it easier to work with.
For Trig subs, you're not relating it back to the inverse functions, those will show up naturally when you do a trig sub. What you're doing is, relating your x^2 and other term back to the 3 important formulas that you learned way way back in trig,\[\large \sin^2x+\cos^2x=1\]\[\large 1+\tan^2x=\sec^2x\]\[\large 1+\cot^2x=\csc^2x\] Err I guess that third one doesn't really matter since it's the same form as the second one :d but whatev.
I mean honestly, it's just a matter of convenience. You can also do hyperbolic trig substitutions. Using: \[\cosh^2(x)-\sinh^2(x)=1\]

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