Why is my answer for integrating tan^3(x) dx wrong?

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SUMMARY

The integral of tan^3(x) dx can be solved using the substitution u = tan(x) and du = sec^2(x) dx. The correct approach involves rewriting tan^3(x) as tan(x)(sec^2(x) - 1) and then separating the integral into two parts: ∫tan(x) sec^2(x) dx and ∫tan(x) dx. The final result includes (1/2)sec^2(x) + ln |cos(x)|, with the solution being correct up to a constant.

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I'm not understanding why my answer is wrong

Homework Statement


\inttan^3(x) dx
This is the solution I've been getting for this problem, but I notice you get a different answer when you let u = tan(x) and du = sec^2(x) dx

Homework Equations


tan^2(x) = (sec^2(x) - 1)

The Attempt at a Solution


\inttan(x) (sec^2(x) - 1) dx
\inttan(x) sec^2(x) - \inttan(x) dx
. // u = sec(x) du = sec(x)tan(x) dx
\intu du - \intsin(x) dx/cos(x)
. // z = cos(x) zu = -sin(x) dx
\intu du - \int-zu/z
(1/2)u^2 - (-ln |z|)
(1/2)sec^2(x) + ln |cos(x)|

Homework Statement


Homework Equations


The Attempt at a Solution

 
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What you have there is correct, up to a constant at least.
 

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