Question:

Let $u = \sec \theta + \tan \theta$. Then $du = (\sec \theta \tan \theta + \sec^2 \theta) d\theta = \sec \theta(\tan \theta + \sec \theta) d\theta = \sec \theta \cdot u \cdot d\theta$. Thus $\sec \theta d\theta = du/u$. Also $\sec \theta = (u + 1/u)/2$ and $\tan \theta = (u - 1/u)/2$. The integral is $\int \sec \theta \cdot u^2 \cdot (du/u) = \int \sec \theta \cdot u \cdot du = \int \frac{u + 1/u}{2} \cdot u \cdot du = \frac{1}{2} \int (u^2 + 1) du = \frac{1}{2} (u^3/3 + u) + C = \frac{u}{6} (u^2 + 3) + C$. Substituting $u = \sec \theta + \tan \theta$, we get $\frac{(\sec \theta + \tan \theta)}{6} ((\sec \theta + \tan \theta)^2 + 3) + C$. This simplifies to option (C).