Integrate f(x)=tan^5(x)/cos^2(x): Get Answer Here

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Discussion Overview

The discussion revolves around the integration of the function f(x) = tan^5(x)/cos^2(x). Participants explore various methods for solving this integral and also touch upon a related integral involving ln(x^2 - 1). The scope includes mathematical reasoning and integration techniques.

Discussion Character

  • Mathematical reasoning
  • Homework-related

Main Points Raised

  • One participant expresses difficulty in integrating f(x) = tan^5(x)/cos^2(x) and suggests that the answer should be tan^6(x)/6 + C.
  • Another participant proposes a substitution method, suggesting u = tan x and notes that du = sec^2 x dx = 1/cos^2 x dx.
  • A different participant seeks assistance with a new integral, ln(x^2 - 1), mentioning they got stuck using integration by parts.
  • One participant suggests using polynomial long division and partial fractions to simplify the integral of 2x^2/(x^2 - 1).
  • Another participant emphasizes the preference for keeping questions within a single thread and discourages starting new questions mid-discussion.
  • A later reply provides a logarithmic identity, stating that ln(x^2 - 1) can be expressed as ln(x + 1) + ln(x - 1).

Areas of Agreement / Disagreement

Participants do not reach a consensus on the integration of f(x) = tan^5(x)/cos^2(x), as the initial claim about the answer remains unverified and alternative methods are suggested. Additionally, the discussion about ln(x^2 - 1) introduces multiple approaches without a clear resolution.

Contextual Notes

Some participants' methods depend on specific assumptions about integration techniques, and there are unresolved steps in the integration processes discussed.

Yankel
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Hello all,

I am trying to integrate the function

f(x)=tan^5(x)/cos^2(x)

I can't figure out how to do it

the answer should be tan^6(x)/6 + C

thanks...
 
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If you make the substitution $u = \tan x$, then $ \displaystyle du = \sec^{2} x \ dx = \frac{1}{\cos^{2} x} \ dx $.
 
Thank you, your tip was great help.

I want to ask for help with one more integral, seems a waste to start a new thread, I need this:

\[ln(x^{2}-1)\]

I tried using integration in parts and got stuck with the integral of:

\[\frac{2x^{2}}{x^{2}-1}\]

Thanks !
 
Using polynomial long division and then the method of partial fractions,

$$\frac{x^{2}}{x^{2}-1} = 1 + \frac{1}{x^{2}-1} = 1 + \frac{1}{2} \Big( \frac{1}{x-1} - \frac{1}{x+1} \Big)$$
 
Yankel said:
I want to ask for help with one more integral, seems a waste to start a new thread,

Actually, that's exactly what we do prefer here at MHB. You can ask two questions per thread, but please do not start a new question in the middle of an existing thread.

Thanks!
 
Yankel said:
I want to ask for help with one more integral, seems a waste to start a new thread, I need this:

\[ln(x^{2}-1)\]

I tried using integration in parts and got stuck with the integral of:

\[\frac{2x^{2}}{x^{2}-1}\]

Thanks !
$$\ln(x^2-1)=\ln(x+1)+\ln(x-1).$$
 

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