$\int \sec ^4(x) ~\mathrm dx$
I did this in two ways:
1)
\begin{align} \int\sec^4(x) ~\mathrm dx & = \int \sec^2(x)\sec^2(x) ~\mathrm dx \\ & = \sec^2(x)\tan(x) - 2\int \sec(x) \tan(x) \sec(x) \tan(x) ~\mathrm dx \\ & = \sec^2(x) \tan(x) - 2 \int \sec^2(x) (\sec^2(x)-1) ~\mathrm dx \\ & = \int \sec^4(x) ~\mathrm dx \\ & = \sec^2(x)\tan(x)-2\int \sec^4(x) ~\mathrm dx +2\int\sec^2(x) ~\mathrm dx \\ & = \int \sec^4(x) ~\mathrm dx \\ & = \frac{\sec^2(x)\tan(x)}{3}+2\tan(x) / 3 +C \end{align}
2) \begin{align} \int \sec^4(x) ~\mathrm dx & = \int (\tan^2(x)+1)^2 ~\mathrm dx \\ & = \int \tan^4(x) + 2\tan^2(x) +1 ~\mathrm dx \\ & = \int \tan^4(x) ~\mathrm dx +2\int \tan^2(x) ~\mathrm dx + x \\ & = (...) \\ & = \frac{\tan^3(x)}{3}-\tan(x) \end{align}
So... which one is wrong( or both?)
