Integral Calculus Question 259
Question: $ \int_{{}}^{{}}{{{(\sec x+\tan x)}^{2}}dx=} $
[MP PET 1987, 92]
Options:
A) $ 2(\sec x+\tan x)-x+c $
B) $ 1/3{{(\sec x+\tan x)}^{3}}+c $
C) $ \sec x(\sec x+\tan x)+c $
D) $ 2(\sec x+\tan x)+c $
Show Answer
Answer:
Correct Answer: A
Solution:
$ \int_{{}}^{{}}{{{(\sec x+\tan x)}^{2}}dx} $ $ =\int_{{}}^{{}}{({{\sec }^{2}}x+{{\tan }^{2}}x+2\sec x\tan x),dx} $ $ =\int_{{}}^{{}}{(2{{\sec }^{2}}x-1+2\sec x\tan x),dx} $ $ =2\tan x+2\sec x-x+c=2(\sec x+\tan x)-x+c. $
BETA
