Integral Calculus Question 359
Question: $ \int_{{}}^{{}}{e^{x}(1+\tan x)\sec x\ dx=} $
[Karnataka CET 2005]
Options:
A) $ e^{x}\cot x $
B) $ e^{x}\tan x $
C) $ e^{x}\sec x $
D) $ e^{x}\cos x $
Show Answer
Answer:
Correct Answer: C
Solution:
$ \int_{{}}^{{}}{e^{x}(1+\tan x)\sec x,dx}=\int_{{}}^{{}}{e^{x}\sec x,dx}+\int_{{}}^{{}}{e^{x}\tan x\sec x,dx} $ $ =e^{x}\sec x-\int_{{}}^{{}}{e^{x}\sec x\tan x,dx}+\int_{{}}^{{}}{e^{x}\sec x\tan x,dx} $ $ =e^{x}\sec x+c. $ Aliter : $ \int_{{}}^{{}}{e^{x}(\sec x+\sec x\tan x),dx}=e^{x}\sec x+c $ Obviously, it is of the form $ \int_{{}}^{{}}{e^{x}{ f(x)+{f}’(x) }},dx. $
BETA
