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The integral of tan^3(x) dx is (1/2)tan^2(x)ln|sec(x)| - (1/2)ln|sec(x)| + C.

To evaluate the integral of tan^3(x) dx, we can use integration by substitution. Let u = tan(x), then du/dx = sec^2(x) and dx = du/sec^2(x). Substituting these into the integral, we get:

∫tan^3(x) dx = ∫u^3(sec^2(x) dx)

= ∫u^3(1 + tan^2(x)) dx

= ∫u^3(1 + u^2) du

= ∫u^3 du + ∫u^5 du

= (1/4)u^4 + (1/6)u^6 + C

Substituting back u = tan(x), we get:

(1/4)tan^4(x) + (1/6)tan^6(x) + C

= (1/2)tan^2(x)(1/2tan^2(x) + ln|sec(x)|) - (1/2)ln|sec(x)| + C

= (1/2)tan^2(x)ln|sec(x)| - (1/2)ln|sec(x)| + C

Therefore, the integral of tan^3(x) dx is (1/2)tan^2(x)ln|sec(x)| - (1/2)ln|sec(x)| + C.

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