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

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SUMMARY

The integral I(tan^3x/cos^4x, x) can be solved using the substitution u = secx, leading to the expression I((u^2-1)u^3, u). The final result is sec^6x/6 - sec^4x/4 + C. An alternative solution presented in the discussion yields tan^6x/6 + tan^4x/4 + C, which is also valid and can be derived from the first solution by applying the identity sec^2x = tan^2x + 1 and simplifying. Both solutions are correct and differ only by a constant term.

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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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