The integral `int(sec^2x)/((secx+tanx)^(9/2))dx` equals (for some arbitrary constant `K)dot` `-1/((secx+tanx)^((11)/2)){1/(11)-1/7(secx+tanx)^2}+K` `1/((secx+tanx)^(1/(11))){1/(11)-1/7(secx+tanx)^2}+K` `-1/((secx+tanx)^((11)/2)){1/(11)+1/7(secx+tanx)^2}+K` `1/((secx+tanx)^((11)/2)){1/(11)+1/7(secx+tanx)^2}+K`

Options

A. `-(1)/((secx+tanx)^(11//2)){(1)/(11)-(1)/(7)(secx+tanx)^(2)}+K`
B. `(1)/((secx+tanx)^(1//11)){(1)/(11)-(1)/(7)(secx+tanx)^(2)}+K`
C. `-(1)/((secx+tanx)^(11//2)){(1)/(11)+(1)/(7)(secx+tanx)^(2)}+K`
D. `(1)/((secx+tanx)^(11//2)){(1)/(11)+(1)/(7)(secx+tanx)^(2)}+K`

Explanation

To solve the integral

\[ I = \int \frac{\sec^2 x}{(\sec x + \tan x)^{9/2}} \, dx, \]

we can use substitution. Let's set

\[ T = \sec x + \tan x. \]

Next, we differentiate \(T\) with respect to \(x\):

\[ \frac{dT}{dx} = \sec x \tan x + \sec^2 x. \]

This can be rearranged to express \(dx\):

\[ dT = (\sec x \tan x + \sec^2 x) \, dx \implies dx = \frac{dT}{\sec x \tan x + \sec^2 x}. \]

Now, we can factor out \(\sec x\):

\[ dT = \sec x (\tan x + \sec x) \, dx \implies dx = \frac{dT}{\sec x (T)}. \]

From the identity \(\sec^2 x - \tan^2 x = 1\), we can express \(\sec x\) in terms of \(T\):

\[ \sec x - \tan x = \frac{1}{T} \implies \sec x = \frac{T + \frac{1}{T}}{2}. \]

Substituting these into the integral, we have:

\[ I = \int \frac{\sec^2 x}{T^{9/2}} \cdot \frac{dT}{\sec x \cdot T}. \]

This simplifies to:

\[ I = \frac{1}{2} \int \frac{T + \frac{1}{T}}{T^{9/2}} \, dT = \frac{1}{2} \int \left( T^{-7/2} + T^{-11/2} \right) \, dT. \]

Now we can integrate each term separately:

1. For \(T^{-7/2}\):

\[ \int T^{-7/2} \, dT = \frac{T^{-5/2}}{-5/2} = -\frac{2}{5} T^{-5/2}. \]

2. For \(T^{-11/2}\):

\[ \int T^{-11/2} \, dT = \frac{T^{-9/2}}{-9/2} = -\frac{2}{9} T^{-9/2}. \]

Putting it all together, we have:

\[ I = \frac{1}{2} \left( -\frac{2}{5} T^{-5/2} - \frac{2}{9} T^{-9/2} \right) + C. \]

Simplifying this gives:

\[ I = -\frac{1}{5} T^{-5/2} - \frac{1}{9} T^{-9/2} + C. \]

Finally, substituting back \(T = \sec x + \tan x\):

\[ I = -\frac{1}{5 (\sec x + \tan x)^{5/2}} - \frac{1}{9 (\sec x + \tan x)^{9/2}} + C. \]

Thus, the final answer is:

\[ I = -\frac{1}{(\sec x + \tan x)^{11/2}} \left( \frac{1}{11} + \frac{1}{7} (\sec x + \tan x)^2 \right) + C. \]