# Simple Trig substitution; find my error

• Feb 6th 2014, 08:17 PM
Tclack
Simple Trig substitution; find my error
So, I was trying to do an easy trig. substitution problems in two different ways (changing which leg to make which on the triangle)
And I came up with two very similar answers, The one that is correct has been determined by both my book and mathematica.

But, I can't see why the wrong one is wrong. Any ideas?
• Feb 7th 2014, 01:05 AM
BobP
Re: Simple Trig substitution; find my error
I'm slightly confused because the answer on the right, the one you say is correct, is missing a negative sign isn't it ?

That aside, the two answers are equivalent, the confusion coming from your use of $\theta$ for both calculations. It would have been better to use $\theta$ for one and $\phi$ for the other.

Then, $\theta + \phi = \pi/2,$ that is, $\tan^{-1}(e^{x}/2)+\tan^{-1}(2/e^{x})=\pi/2,$ or

$\tan^{-1}(e^{x}/2)=\pi/2-\tan^{-1}(2/e^{x}).$

The $\pi/2$ gets swallowed up by the constant of integration.
• Feb 7th 2014, 09:19 AM
Tclack
Re: Simple Trig substitution; find my error
You are correct, There IS a missing negative. Good catch. BUT, that aside:

These are independent triangles (the one on the right is not the one on the left flipped over with me evaluating the other angle);
So the dilemma is that I could have chosen either leg to be 2 and either to be e^x, why do I get two different answers?

I actually just found another post in the trigonometry section with the [I]same[I] type of problem. Here:
http://mathhelpforum.com/trigonometr...tml#post809986

This must be a common occurrence.
• Feb 8th 2014, 12:13 AM
BobP
Re: Simple Trig substitution; find my error
Both results are correct (after that negative sign has been included in the right hand one).
They are just two different ways of expressing the same result.
Check it by evaluating both results between limits.
• Feb 8th 2014, 07:29 AM
Tclack
Re: Simple Trig substitution; find my error
I don't know what you mean by evaluating between limits. Are you talking about the squeeze theorem? It seems much easier to evaluate by setting them equal to eachother, which is giving an erroneous result:
Attachment 30144
So the error either lies in the check I did above or in the setup or evaluation of the problem (as shown in my first picture posted....except for that stupid negative sign mistake)
Aggh, this is driving me crazy. If I can't see what I'm doing wrong, it takes away all the confidence I have in the results of all other trig. substitution problems.
• Feb 8th 2014, 08:06 AM
BobP
Re: Simple Trig substitution; find my error
Look again at the result towards the end of #2, that is the relationship between the two arctangents, (not the one in your last post which is wrong).
The $\pi/2$ doesn't appear in the result for the second of the integrals because it is taken up by the constant of integration.
By checking between limits, I meant choose some random limits of integration and evaluate the integral according to your two results, they will be the same.
• Feb 8th 2014, 04:13 PM
Tclack
Re: Simple Trig substitution; find my error
I'm sorry Bob, I'm not understanding what you are saying above. Could you be more clear or show me exactly how the two answers are equivalent if indeed they are? (I really do appreciate your effort). Moreover, what's wrong with the second picture? Both angles are the same, so they should produce a true statement when I set them equal to each other. (i.e. Both $\theta$'s ARE the same angle $\theta$; that's how I set them up.)

(I can't emphasize it enough, I'm evaluating the SAME angle in two different ways. I DO know that if one angle in a triangle is $\pi/2$, then the sum of the other two angles is also $\pi/2$. But that's not the case here. So, it shouldn't matter which leg I choose to be $e^x$ and which one I choose to be $2$. )
• Feb 9th 2014, 01:18 AM
BobP
Re: Simple Trig substitution; find my error
Forget your triangles and angles, and look again at the relationship given at the bottom of #2, (the one that links the two arctangents).

Convince yourself that it is correct. You can do that by choosing (several times) some arbitrary value for x and substituting in on both sides of the equation, and seeing that both sides produce the same result. Alternatively, if you have access to a graph plotter, plot each side separately, the two graphs will be the same.

Now look at the second of the two results, the one that you label as correct, (but remember to put that negative sign in). Substitute using the relationship discussed above.
-arctan(2/exp(x))=arctan(exp(x)/2) - pi/2, (sorry, latex isn't working for me).

The pi/2 gets lumped together with the existing constant of integration, constant plus constant equals constant.
You finish up with your first result, (which is simply an alternative form for the result and is as equally correct as the other).
• Feb 10th 2014, 09:07 AM
Tclack
Re: Simple Trig substitution; find my error
I understand now. I took the derivative of both sides on the first picture (after the negative was put in) and got back to the original problem. Thank you for your patience Bob.