Indefinite Integral of Secant and Tangent Algebraic Expression

To evaluate the indefinite integral $I = \int \frac{\sec^2 x}{(\sec x + \tan x)^{9/2}} \, dx$

we perform a change of variables.

Let $t = \sec x + \tan x$

Differentiating with respect to $x$: $\frac{dt}{dx} = \sec x \tan x + \sec^2 x = \sec x (\sec x + \tan x) = t \sec x$ which gives: $\sec x \, dx = \frac{dt}{t}$

We also know the fundamental trigonometric identity: $\sec^2 x - \tan^2 x = 1 \implies (\sec x - \tan x)(\sec x + \tan x) = 1$ Hence, $\sec x - \tan x = \frac{1}{t}$

Adding the expressions for $t$ and $\frac{1}{t}$: $(\sec x + \tan x) + (\sec x - \tan x) = t + \frac{1}{t}$ $2 \sec x = t + \frac{1}{t} \implies \sec x = \frac{1}{2} \left( t + \frac{1}{t} \right)$

Now, rewriting the integrand in terms of $t$: $I = \int \frac{\sec x \cdot (\sec x \, dx)}{(\sec x + \tan x)^{9/2}} = \int \frac{\frac{1}{2} \left( t + \frac{1}{t} \right) \cdot \frac{dt}{t}}{t^{9/2}}$

Simplifying the algebraic expression: $I = \frac{1}{2} \int \frac{t + t^{-1}}{t^{11/2}} \, dt = \frac{1}{2} \int \left( t^{-9/2} + t^{-13/2} \right) dt$

Integrating term by term: $\int t^{-9/2} \, dt = \frac{t^{-9/2 + 1}}{-9/2 + 1} = \frac{t^{-7/2}}{-7/2} = -\frac{2}{7} t^{-7/2}$ $\int t^{-13/2} \, dt = \frac{t^{-13/2 + 1}}{-13/2 + 1} = \frac{t^{-11/2}}{-11/2} = -\frac{2}{11} t^{-11/2}$

Substituting these back into $I$: $I = \frac{1}{2} \left[ -\frac{2}{7} t^{-7/2} - \frac{2}{11} t^{-11/2} \right] + K$ $I = -\left[ \frac{1}{7} t^{-7/2} + \frac{1}{11} t^{-11/2} \right] + K$

Factoring out $-\frac{1}{t^{11/2}}$: $I = -\frac{1}{t^{11/2}} \left\{ \frac{1}{11} + \frac{1}{7} t^2 \right\} + K$

Substituting back $t = \sec x + \tan x$: $I = -\frac{1}{(\sec x + \tan x)^{11/2}} \left\{ \frac{1}{11} + \frac{1}{7}(\sec x + \tan x)^2 \right\} + K$

This corresponds to Option (C).