The Question:

$\displaystyle \int {sec^4\theta \ d\theta}$

My Attempt:

$\displaystyle \int {sec^2\theta \ sec^2\theta \ d\theta}$

$\displaystyle \int {sec^2\theta(tan^2\theta + 1) \ d\theta}$

Let $\displaystyle u = tan\theta$

$\displaystyle du = sec^2\theta \ d\theta$

$\displaystyle \int {u^2 + 1 \ du}$

= $\displaystyle \frac{1}{3}u^3 + u$

= $\displaystyle \frac{1}{3}tan^3\theta + tan\theta$

However, according to Wolfram Alpha the solution is this.

What am I doing incorrectly? Thanks.