The value of ^3 x , ,dx will be - Vidyadip

MCQ

The value of $\int {{{\sec }^3}x\,\,dx} $ will be

✓
$\frac{1}{2}\left[ {\,\sec x\tan x + \log (\sec x + \tan x)} \right]$
B
$\frac{1}{3}\left[ {\,\sec x\tan x + \log (\sec x + \tan x)} \right]$
C
$\frac{1}{4}\left[ {\,\sec x\tan x + \log (\sec x + \tan x)} \right]$
D
$\frac{1}{8}\left[ {\,\sec x\tan x + \log (\sec x + \tan x)} \right]$

✓

Answer

Correct option: A.

$\frac{1}{2}\left[ {\,\sec x\tan x + \log (\sec x + \tan x)} \right]$

a
(a) Let $I = \int {{{\sec }^3}xdx} $$ = \int {\sec x{{\sec }^2}xdx} $
$ \Rightarrow I = \sec x\tan x - \int {\sec x\,{{\tan }^2}x\,dx} $
$ \Rightarrow I = \sec x\tan x - \int {\sec x\,({{\sec }^2}x - 1)dx} $
$ \Rightarrow I = \sec x\tan x - \int {{{\sec }^3}x\,dx} + \int {\sec x\,dx} $
$ \Rightarrow \;I = \sec x\tan x - I + \log \,(\sec x\, + \tan x\,)$
$ \Rightarrow 2I = \sec x\tan x + \log (\sec x + \tan x)$
$ \Rightarrow I = \frac{1}{2}[\sec x\tan x + \log (\sec x + \tan x)]$.

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