# FP3 IntegrationWatch

#1
Trying to integrate . My book, and Wolfram give the answer , using a hyperbolic substitution, but I get the answer using a trigonometric substitution (). Am I wrong, or are there just multiple ways of doing this?
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7 years ago
#2
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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#3
They did ask for a substitution, but didn't specify whether it had to be hyperbolic or trigonometric.
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7 years ago
#4
Neither, in this case it would be to let u=x+2.
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7 years ago
#5
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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7 years ago
#6
(Original post by ViralRiver)
Am I wrong, or are there just multiple ways of doing this?
7 years ago
#7
(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 . My book, and Wolfram give the answer , using a hyperbolic substitution, but I get the answer using a trigonometric substitution (). 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 . Therefore the answer , which is not what you got for the logarithm version.
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#8
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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7 years ago
#9
If that's the case, could it be using a subsitution of say, ?

Edit: Oh, and for the record the substitution you have chosen to use works too, but as farhan says you haven't expressed in terms of correctly. Once you have done this the logarithmic equivalent of will equal your simplified expression.
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7 years ago
#10
How did you get from to ? Note that . You should get the same answer as in the book.

To be honest, if you see then the substitution is usually safer than , which is better for just on its own (or with an integer power).
#11
I used the substitution . This reduced the integrand to , giving . I used 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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7 years ago
#12
(Original post by ViralRiver)
I used the substitution . This reduced the integrand to , giving . I used 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 >< .
. See my previous post.
#13
Ahh, as I said, stupid mistake xD . Not sure what I was thinking, but this is resolved.
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