Integration involving trig. substitutionby Mangoes Tags: integration, involving, substitution, trig 

#1
Jul3112, 08:00 PM

P: 71

1. The problem statement, all variables and given/known data
See below. 2. The attempt at a solution Hey there, apologies for doing so, but I don't know how to use latex on this board (it doesn't show me anything when I hit preview, so I had to make a picture for the sake of cleanness. imgur.com/WzrAR.png The first line is the original problem, with the proceeding lines being the attempt I made. I have about four months before I transfer to a new university and figured I might as well selfstudy Calculus. I finished Calculus I without any trouble so I just started doing what I will eventually cover in Calculus II, but unfortunately my textbook offers only a mechanical way of doing this technique which leaves some doubt in my mind. Is trigonometric substitution legal to do in any integral (regardless of whether it will actually help integrating) so long as you also substitute any related variables to the one you're substituting with? I've tried it out with simple indefinite integrals like ∫3x^2 dx but I'm unsure whether or not issues would occur if I substituted, for example, x = tan(θ) for ∫1/x due to the fact that 1/tan(θ) is undefined in an infinite amount of inputs while 1/x is undefined in only one. Furthermore, I've looked over my work and I can't find any errors, so I don't know whether there's some concept I'm completely missing out on or there's some hidden error I just can't see. Thanks for any input. 



#2
Jul3112, 08:19 PM

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P: 25,170

You are messing up in the integration by parts section. Otherwise, well done! You are partially ignoring the presence of the 9/4 factor. Try it this way. Just integrate sec^3 by parts. Add the 9/4 factor later.




#3
Jul3112, 08:29 PM

Mentor
P: 21,012

Your work looks good, and nothing immediately pops out at me. I think that you have a constant wrong somewhere.
I'll take a closer look and see if I can spot where the problem is. I suppose trig substitution could be applied to any integral, but where it's most commonly used is where you have the square root of a sum or difference of squares. The radical can appear in the numerator, but commonly appears in the denominator. The idea with trig substitution is to apply the substitution to get an integral that you can actually evaluate, so I don't see it used with cube roots, or the like, or with simple polynomials or power functions, as in your example. LaTeX isn't hard to use. Here's the LaTeX for the integral you started with. To make it actually render, put a pair of $ signs at the beginning, and another pair at the end. \int \sqrt{9 + 16x^2}dx Edit: Dick beat me to it, so I won't take a closer look. 



#4
Jul3112, 09:08 PM

P: 71

Integration involving trig. substitutionHowever, I worked it out for a specific example on paper and saw that when θ gave undefined values, the corresponding x for that θ would have given undefined values anyways. Sorry if my wording was a little strange, but I'm satisfied now. It had been bothering me for a couple of days. 



#5
Jul3112, 10:01 PM

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It should all work out ok. If you take your original example of integrating 1/x by substituting x=tan(t) you should wind up with log(tan(t)). Sure that has problems all over the place, but when you change back to x you get log(x) and everything is ok again. Glad you are feeling better about it.



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