Question:

Step 1: Define the substitution and find its differential. Let the given integral be $I$. We use the substitution $u = \sec \theta + \tan \theta$. To find $du$, we differentiate $u$ with respect to $\theta$: $$ \frac{du}{d\theta} = \frac{d}{d\theta}(\sec \theta + \tan \theta) = \sec \theta \tan \theta + \sec^2 \theta $$ Factor out $\sec \theta$: $$ \frac{du}{d\theta} = \sec \theta (\tan \theta + \sec \theta) $$ Substitute $u = \sec \theta + \tan \theta$ back into the expression for $du/d\theta$: $$ \frac{du}{d\theta} = \sec \theta \cdot u $$ From this, we can express $\sec \theta d\theta$ in terms of $u$ and $du$: $$ \sec \theta d\theta = \frac{du}{u} $$ Step 2: Express $\sec \theta$ in terms of $u$. We have $u = \sec \theta + \tan \theta$. We also know the identity $\sec^2 \theta - \tan^2 \theta = 1$. Factoring the left side gives $(\sec \theta - \tan \theta)(\sec \theta + \tan \theta) = 1$. Substituting $u$: $$ (\sec \theta - \tan \theta)u = 1 $$ $$ \sec \theta - \tan \theta = \frac{1}{u} $$ Now we have a system of two equations: 1. $u = \sec \theta + \tan \theta$ 2. $\frac{1}{u} = \sec \theta - \tan \theta$ Adding the two equations: $$ u + \frac{1}{u} = (\sec \theta + \tan \theta) + (\sec \theta - \tan \theta) $$ $$ u + \frac{1}{u} = 2 \sec \theta $$ Therefore, $\sec \theta = \frac{1}{2} \left(u + \frac{1}{u}\right)$. Step 3: Rewrite the original integral in terms of $u$ and $du$. The given integral is $I = \int \sec^2 \theta (\sec \theta + \tan \theta)^2 d\theta$. We can split $\sec^2 \theta$ into $\sec \theta \cdot \sec \theta$: $$ I = \int \sec \theta \cdot (\sec \theta + \tan \theta)^2 \cdot (\sec \theta d\theta) $$ Now, substitute $u = \sec \theta + \tan \theta$ and $\sec \theta d\theta = \frac{du}{u}$: $$ I = \int \sec \theta \cdot u^2 \cdot \frac{du}{u} $$ Simplify the expression: $$ I = \int \sec \theta \cdot u \cdot du $$ Step 4: Substitute the expression for $\sec \theta$ in terms of $u$ into the integral. Substitute $\sec \theta = \frac{1}{2} \left(u + \frac{1}{u}\right)$ into the integral: $$ I = \int \frac{1}{2} \left(u + \frac{1}{u}\right) \cdot u \cdot du $$ $$ I = \frac{1}{2} \int \left(u^2 + 1\right) du $$ Step 5: Perform the integration. Integrate the polynomial term by term: $$ I = \frac{1}{2} \left(\frac{u^3}{3} + u\right) + C $$ $$ I = \frac{u^3}{6} + \frac{u}{2} + C $$ Factor out $\frac{u}{6}$: $$ I = \frac{u}{6} (u^2 + 3) + C $$ Step 6: Substitute back to express the result in terms of $\theta$. Substitute $u = \sec \theta + \tan \theta$ back into the expression: $$ I = \frac{(\sec \theta + \tan \theta)}{6} ((\sec \theta + \tan \theta)^2 + 3) + C $$ Step 7: Simplify the expression to match one of the given options. Let's expand the term $(\sec \theta + \tan \theta)^2 + 3$: $$ (\sec \theta + \tan \theta)^2 + 3 = (\sec^2 \theta + \tan^2 \theta + 2 \sec \theta \tan \theta) + 3 $$ Using the identity $\sec^2 \theta = 1 + \tan^2 \theta$: $$ = (1 + \tan^2 \theta + \tan^2 \theta + 2 \sec \theta \tan \theta) + 3 $$ $$ = 1 + 2 \tan^2 \theta + 2 \sec \theta \tan \theta + 3 $$ $$ = 4 + 2 \tan^2 \theta + 2 \sec \theta \tan \theta $$ Factor out 2: $$ = 2 (2 + \tan^2 \theta + \sec \theta \tan \theta) $$ Rearrange the terms: $$ = 2 (2 + \tan \theta (\tan \theta + \sec \theta)) $$ Now substitute this back into the integral expression: $$ I = \frac{(\sec \theta + \tan \theta)}{6} \cdot 2 (2 + \tan \theta (\tan \theta + \sec \theta)) + C $$ $$ I = \frac{(\sec \theta + \tan \theta)}{3} (2 + \tan \theta (\sec \theta + \tan \theta)) + C $$ This matches Option (C). The final answer is $\boxed{\frac{(\sec \theta + \tan \theta)}{3} [2 + \tan \theta(\sec \theta + \tan \theta)] + C}$.