MCQ
$\int_{}^{} {{{\{ 1 + 2\tan x(\tan x + \sec x)\} }^{1/2}}dx = } $
- A$\log (\sec x + \tan x) + c$
- B$\log {(\sec x + \tan x)^{1/2}} + c$
- ✓$\log \sec x(\sec x + \tan x) + c$
- DNone of these
✓
Answer
Correct option: C.
$\log \sec x(\sec x + \tan x) + c$
c
(c)$\int_{}^{} {{{(1 + 2{{\tan }^2}x + 2\tan x\sec x)}^{1/2}}dx} $
$ = \int_{}^{} {{{({{\sec }^2}x + {{\tan }^2}x + 2\tan x\sec x)}^{1/2}}dx} $
$ = \int_{}^{} {(\sec x + \tan x)\,dx} = \log (\sec x + \tan x) + \log \sec x + c$
$ = \log \sec x(\sec x + \tan x) + c$.
(c)$\int_{}^{} {{{(1 + 2{{\tan }^2}x + 2\tan x\sec x)}^{1/2}}dx} $
$ = \int_{}^{} {{{({{\sec }^2}x + {{\tan }^2}x + 2\tan x\sec x)}^{1/2}}dx} $
$ = \int_{}^{} {(\sec x + \tan x)\,dx} = \log (\sec x + \tan x) + \log \sec x + c$
$ = \log \sec x(\sec x + \tan x) + c$.
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