How do I find the antiderivative of -tanx on the interval (-π/2, π/2)?

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Homework Help Overview

The discussion revolves around finding the antiderivative of the function f(x) = -tan(x) on the interval (-π/2, π/2). Participants are exploring integration techniques as part of their learning process in calculus.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss various methods for integration, including substitution and manipulation of the function. There is uncertainty about how to approach the problem, particularly regarding the relationship between differentiation and integration of trigonometric functions.

Discussion Status

Some participants have offered suggestions for methods, such as multiplying by (sec(x) + tan(x)) and using substitution. However, there is a recognition that the problem may be challenging for those new to integration, and not all participants feel confident in their understanding of the techniques discussed.

Contextual Notes

One participant notes that the problem is categorized as a "challenger" question, indicating it may be more advanced than typical introductory problems. There is also mention of the difficulty in finding antiderivatives for certain functions, suggesting that some functions may not have straightforward methods for integration.

LHC
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Hi, I'm just starting to learn integration (I went to the first class in my summer course yesterday), and I'm already a bit confused. I was doing some practice problems in the text, and I found this:

Find the most general antiderivative of f on the indicated interval:

f(x) = -\tan x on (-\frac{\pi}{2}, \frac{\pi}{2})

I'm...really unsure as to how to do this problem. All this time I'm thinking, "What do I differentiate to get tan x in the first place?" ...heh, I'm still not sure. Could someone please give me some pointers? Thanks!
 
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Try multiplying by (secx + tanx)/(secx + tanx).
 
-\int\frac{\sin x}{\cos x}dx

u=\cos x \rightarrow -du=\sin xdx
 
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I took your suggestion and worked it out on paper, but...I don't seem to be able to see how this would help.

f(x) = (-sec x tan x - tan^2 x)/(sec x + tan x)...

I'm afraid I've run into something that's a bit too advanced (this problem was listed under the "challenger" questions..and I was just curious). I will come back to this problem once I'm more...knowledgeable? Nevertheless, thank you for your help. I shall ask my teacher to explain it to me in detail later.
 
The method used to solved that is called integration by substitution. Basically, you make a part of the equation equal to u (I'd choose cos x) and find du (which is equal to -sinx*dx). Substitute in and integrate in terms of u.
 
durt said:
Try multiplying by (secx + tanx)/(secx + tanx).

I believe you're thinking for the trick for integrating sec x ...

You may be having trouble with this one because it's not generally a function students are asked to find the antiderivative of before they have seen the basic antiderivatives. And tan x is not the derivative of any of the elementary functions. Were you assigned this problem for practice or did you pick it out yourself? I ask because you will quickly discover that, while after having learned the rules of differentiation, you can find the derivative of any messy function you can cook up out of the elementary functions, it is rather easy to pick out or make up a function that is extremely difficult to anti-differentiate (and there are some no one has found a method for: these are integrated numerically).

As some of the posters have mentioned already, the trick here is to exploit the fact that the derivative of ln(x) is 1/x , so we can say d[ ln(x) ] = 1/x dx = dx / x . This gives us a method for taking anything that can be write as

( derivative of u ) / u ,

and expressing its general antiderivative as ln |u| + C . You will see this used a lot in your course before too much longer...
 
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