What mistake did I make in solving the integral of tan(x)^3?

[Goldenwind]

[SOLVED] Integral of tan(x)^3

Homework Statement

$$\int tan^3 x dx$$

The Attempt at a Solution

$$= \int tanx*tan^2x dx$$
$$= \int tanx*(1 - sec^2x) dx$$
$$= \int tanx - tanxsec^2x dx$$
$$= \int tanx dx - \int tanxsec^2x dx$$
$$= \int \frac{sinx}{cosx} dx - \int u du$$
$$= -\int \frac{-sinx}{cosx} dx - \frac{1}{2}u^2$$
$$= -\int \frac{D(cosx)}{cosx} dx - \frac{1}{2}tanx^2$$
$$= -ln(cosx) - \frac{1}{2}tanx^2 + C$$
$$= ln(cosx^{-1}) - \frac{1}{2}tanx^2 + C$$
$$= ln(\frac{1}{cosx}) - \frac{1}{2}tanx^2 + C$$
$$= ln(secx) - \frac{1}{2}tanx^2 + C$$

Book says I'm wrong. Where is my mistake?

 

[Mathdope]

1 + tan^2 = sec^2 is not equivalent to 1 - sec^2 = tan^2? (step 2) Looks like you missed a negative sign, pretty small error that apparently got magnified later on.

 

[Goldenwind]

...wow... damnit.
lol, thank-you.

 

[shelovesmath]

I'm wondering if this method is also legal.

 

[Quinzio]

I'm wondering if this method is also legal.

yes sure

 

[shelovesmath]

yes sure

Ok, but my answer is different. . .