How do I solve this trigonometric integral: I(tan^3x/cos^4x, x)?

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Homework Help Overview

The discussion revolves around solving a trigonometric integral involving the expression I(tan^3x/cos^4x, x). Participants are exploring different forms of the integral and comparing solutions derived from various approaches.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss different representations of the integral, including transformations using secant and tangent functions. Questions arise regarding the correctness and equivalence of the solutions presented, particularly in relation to constants of integration.

Discussion Status

The discussion is active, with participants validating each other's approaches and exploring the relationships between different forms of the integral. Some guidance is provided on how to derive one solution from another, indicating a productive exchange of ideas.

Contextual Notes

There is mention of discrepancies between the original poster's solution and that found in their textbook, which raises questions about the nature of the solutions and whether they differ beyond the constant of integration.

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Homework Statement




I(tan^3x/cos^4x,x)
I(tan^3x * sec^4x,x)
I((sec^2x-1)sec^4x*tanx,x)
u=secx du=secxtanx
I((u^2-1)u^3,u)
u^6/6-u^4/4+C
sec^6x/6-sec^4x/4+C

im new to this and my book is showing diffrent solutions i see nothing wrong here

Homework Equations





The Attempt at a Solution

 
Last edited:
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Your answer is correct. What is the solution from your book?
 
tan^6x/6+tan^4x/4+C
 
Are you sure the two solutions are different? (By more than just adding a constant.)
 
That answer is also correct. You can get that answer from yours by using sec2 = tan2x + 1, multiplying out the numerators, and simplifying.

The answer you got seems to be the easiest integral to find for the problem.
 

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