How do I solve $\int \sec^3 \theta d\theta$ 
Possible Duplicate:
Indefinite integral of secant cubed 

I guess I need to learn a new technique because those I know didn't help me here.  Wolframalpha uses the reduction formula, which I haven't been introduced to yet.  So is there a way to solve this without using the reduction formula?
 A: I use integration by parts and $\tan^2\theta = \sec^2\theta - 1$.
\begin{align*}
\int\sec^3\theta\ d\theta
& = \int\sec\theta\ d(\tan\theta) \\
& = \sec\theta\tan\theta - \int \tan\theta\ d(\sec\theta) \\
& = \sec\theta\tan\theta - \int \sec\theta\tan^2\theta\ d\theta \\
& = \sec\theta\tan\theta - \int \sec\theta(\sec^2\theta - 1)\ d\theta \\
& = \sec\theta\tan\theta + \int\sec\theta\ d\theta - \int \sec^3\theta\ d\theta \\
\therefore
2 \int\sec^3\theta\ d\theta & =
\sec\theta\tan\theta + \int\sec\theta\ d\theta.
\end{align*}
A: Yes; it involves a couple of integrations by parts and some manipulation with trig identities.
Let $I=\int\sec^3\theta~d\theta$. Let $u=\sec\theta$ and $dv=\sec^2\theta~d\theta$; then $du=\sec\theta\tan\theta~d\theta$ and $v=\tan\theta$, so
$$I=\sec\theta\tan\theta-\int\sec\theta\tan^2\theta~d\theta\;.\tag{1}$$
Now $\tan^2\theta=\sec^2\theta-1$, so
$$\int\sec\theta\tan^2\theta~d\theta=\int\left(\sec^3\theta-\sec\theta\right)d\theta=I-\int\sec\theta~d\theta\;.\tag{2}$$
Combining $(1)$ and $(2)$ yields
$$I=\sec\theta\tan\theta-\left(I-\int\sec\theta~d\theta\right)=\sec\theta\tan\theta-I+\int\sec\theta~d\theta\;,$$ so
$$I=\frac12\left(\sec\theta\tan\theta+\int\sec\theta~d\theta\right)\;.$$
But $$\int\sec\theta~d\theta=\int\frac{\sec\theta(\sec\theta+\tan\theta)}{\sec\theta+\tan\theta}d\theta=\int\frac{d(\sec\theta+\tan\theta)}{\sec\theta+\tan\theta}=\ln(\sec\theta+\tan\theta)+C\;,$$ so finally
$$I=\frac12\sec\theta\tan\theta+\frac12\ln(\sec\theta+\tan\theta)+C\;.$$
A: Make the change of variables $ \sec(x)^2 = 1-u $, which implies
$$ \int \sec(\theta)^3 d \theta = -\frac{1}{2}\int \frac{\sqrt{u-1}}{\sqrt{u}} du \,. $$
Following it with the change of variables $ u = \sin(t)^2 $ yields
$$ -\frac{1}{2}\int \frac{\sqrt{u-1}}{\sqrt{u}} du  = -i \int \cos(t)^2\,dz= -i \left(\frac{1}{4} \sin \left( 2 t \right) + \frac{t}{2} \right)\,. $$
Substituting $ t = \arcsin{\sqrt{u}} $ and then $ u = -\tan(x)^2 $ gives the desired result.  
