The integral ∫sec2⁡x(sec⁡x+tan⁡x)9/2dx equals (for some arbitrary constant K)

A
$- \frac{1}{(\sec x + \tan x)^{11 / 2}} \{\frac{1}{11} - \frac{1}{7} (\sec x + \tan x)^{2}\} + K$
B
$\frac{1}{(\sec x + \tan x)^{11 / 2}} \{\frac{1}{11} - \frac{1}{7} (\sec x + \tan x)^{2}\} + K$
C
$- \frac{1}{(\sec x + \tan x)^{11 / 2}} \{\frac{1}{11} + \frac{1}{7} (\sec x + \tan x)^{2}\} + K$
D
$\frac{1}{(\sec x + \tan x)^{11 / 2}} \{\frac{1}{11} + \frac{1}{7} (\sec x + \tan x)^{2}\} + K$

Solution:

We have,

$I = \int \frac{\sec^{2} x}{(\sec x + \tan x)^{9 / 2}} d x$

Let $s e c x + \tan x = t$ .

. Then, $\sec x - \tan x = 1 / t$ and,

$\sec x (\sec x + \tan x) d x = d t$

$∴ scc x d x = \frac{1}{t} d t$ and, $\sec x = \frac{1}{2} (t + \frac{1}{t})$

$∴ I = \frac{1}{2} \int \frac{\frac{1}{t} (t + \frac{1}{t})}{t^{9 / 2}} d t = \frac{1}{2} \int \frac{1}{t^{9 / 2}} + \frac{1}{t^{13 / 2}} d t$

$\begin{array}{l} \Rightarrow I = - \frac{1}{7 t^{7 / 2}} - \frac{1}{11 t^{11 / 2}} + k \\ \Rightarrow I = - \frac{1}{t^{11 / 2}} \{\frac{t^{2}}{7} + \frac{1}{11}\} + k \\ \Rightarrow I = - \frac{1}{(\sec x + \tan x)^{11 / 2}} \{\frac{1}{11} + \frac{1}{7} (\sec x + \tan x)^{2}\} + k \end{array}$

The integral $\int \frac{\sec^{2} x}{(\sec x + \tan x)^{9 / 2}} d x$ equals (for some arbitrary constant K)

Solution:

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