$If P = \int \frac{\sec^{2} x}{(\sec x + \tan x)^{10}} d x then P is equal to$

Moderate

A

$\frac{1}{18 {[\sec x + \tan x]}^{9}} - \frac{1}{22 {[\sec x + \tan x]}^{11}} + K$
B

$\frac{1}{18 {[\sec x + \tan x]}^{9}} + \frac{1}{11 {[\sec x + \tan x]}^{11}} + K$
C

$\frac{- 1}{18 {[\sec x + \tan x]}^{9}} - \frac{1}{22 {[\sec x + \tan x]}^{11}} + K$
D

$\frac{- 1}{18 {[\sec x + \tan x]}^{9}} + \frac{1}{22 {[\sec x + \tan x]}^{11}} + K$

Solution

$Let \sec x + \tan x = t$
$\begin{array}{l} \Rightarrow (\sec x \tan x + \sec^{2} x) d x = d t \\ \Rightarrow \sec x d x = \frac{d t}{t} \end{array}$
$And \sec x + \tan x = t \to (1)$
$\begin{aligned} \sec x - \tan x = \frac{1}{t} \to (2) \\ (1) + (2) 2 \sec x = t + \frac{1}{t} \end{aligned}$
$\begin{array}{l} \sec x = \frac{1}{2} (\frac{1 + t^{2}}{t}) \\ ∴ P = \int \frac{\frac{1}{2} (\frac{t^{2} + 1}{t}) \frac{d t}{t}}{t^{10}} \\ = \frac{1}{2} \int \frac{t^{2} + 1}{t^{12}} d t \\ = \frac{1}{2} \int (\frac{1}{t^{10}} + \frac{1}{t^{12}}) d t \\ = \frac{1}{2} [\frac{- 1}{9 t^{9}} - \frac{1}{11 t^{11}}] + K \\ = \frac{- 1}{18 [\sec x + \tan x] 9} - \frac{1}{22 [\sec x + \tan x]^{11}} + K \end{array}$

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