FP3 Integration Watch

ViralRiver
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#1
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#1
Trying to integrate \frac{1}{\sqrt{x^2+4x+5}}. My book, and Wolfram give the answer arsinh(x+2), using a hyperbolic substitution, but I get the answer ln|sec\theta +tan\theta|=ln|x^2+5x+7| using a trigonometric substitution (x + 2 = tan\theta). Am I wrong, or are there just multiple ways of doing this?
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olipal
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#2
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Hmm, I guess for a question like this I would probably play it safe and just complete the square in the denominator. Unless they specifically ask for a substitution, that's normally the best move.
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ViralRiver
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#3
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They did ask for a substitution, but didn't specify whether it had to be hyperbolic or trigonometric.
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olipal
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#4
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Neither, in this case it would be to let u=x+2.
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mir3a
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#5
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Complete the square, and let u=x+2. Why use a hyperbolic substitution? All you're doing is making the process much longer.
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ghostwalker
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#6
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(Original post by ViralRiver)
Am I wrong, or are there just multiple ways of doing this?
Differentiate your answer and see what you get.
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Farhan.Hanif93
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#7
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(Original post by olipal)
Hmm, I guess for a question like this I would probably play it safe and just complete the square in the denominator. Unless they specifically ask for a substitution, that's normally the best move.
(Original post by mir3a)
Complete the square, and let u=x+2. Why use a hyperbolic substitution? All you're doing is making the process much longer.
That's assuming he has a formula book to hand or that he remembers the standard integrals. Which may or may not be the case.

(Original post by ViralRiver)
Trying to integrate \frac{1}{\sqrt{x^2+4x+5}}. My book, and Wolfram give the answer arsinh(x+2), using a hyperbolic substitution, but I get the answer ln|sec\theta +tan\theta|=ln|x^2+5x+7| using a trigonometric substitution (x + 2 = tan\theta). Am I wrong, or are there just multiple ways of doing this?
Both methods are fine but I think your logarithmic answer is wrong. To check if they're equivalent, note that \sinh ^{-1}x = \ln (x+\sqrt{x^2+1}). Therefore the answer \sinh ^{-1}(x+2) = \ln ((x+2)+\sqrt{x^2+4x+5}), which is not what you got for the logarithm version.
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ViralRiver
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#8
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Not sure you're all understanding. The question asked me to use a substitution, hyperbolic or trigonometric. It didn't specify which, but it doesn't allow for a normal algebraic substitution. Farhan, I went through that process myself and realised they weren't the same, but had a feeling it may have been something to do with the constant?
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olipal
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#9
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If that's the case, could it be using a subsitution of say, x+2=\sinh u?

Edit: Oh, and for the record the substitution you have chosen to use works too, but as farhan says you haven't expressed \sec \theta in terms of \tan \theta correctly. Once you have done this the logarithmic equivalent of arsinh (x+2) will equal your simplified expression.
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nuodai
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#10
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#10
How did you get from \ln |\sec \theta + \tan \theta| to \ln |x^2+5x+7|? Note that \sec \theta = \sqrt{1+\tan^2 \theta}. You should get the same answer as in the book.

To be honest, if you see \sqrt{1+x^2} then the substitution x=\sinh u is usually safer than x=\tan \theta, which is better for just 1+x^2 on its own (or with an integer power).
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ViralRiver
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#11
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I used the substitution x+2=tan\theta. This reduced the integrand to sec\theta, giving ln|sec\theta +tan\theta|. I used sec\theta =1+tan^2\theta =x^2+4x+5 and substituted it into the logarithm. I can't see anything wrong with it, but I'm sure there's something simple I'm missing, or I integrated incorrectly >< .
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nuodai
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#12
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#12
(Original post by ViralRiver)
I used the substitution x+2=tan\theta. This reduced the integrand to sec\theta, giving ln|sec\theta +tan\theta|. I used sec\theta =1+tan^2\theta =x^2+4x+5 and substituted it into the logarithm. I can't see anything wrong with it, but I'm sure there's something simple I'm missing, or I integrated incorrectly >< .
\sec \theta \not \equiv 1 + \tan^2 \theta. See my previous post.
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ViralRiver
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#13
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Ahh, as I said, stupid mistake xD . Not sure what I was thinking, but this is resolved.
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