Integral Calculus Question 363
Question: $ \int_{{}}^{{}}{e^{x}
[\tan x-\log (\cos x)]\ dx=} $ [MP PET 1991]
Options:
A) $ e^{x}\log (\sec x)+c $
B) $ e^{x}\log (\cos ecx)+c $
C) $ e^{x}\log (\cos x)+c $
D) $ e^{x}\log (\sin x)+c $
Show Answer
Answer:
Correct Answer: A
Solution:
$ \int_{{}}^{{}}{e^{x}[\tan x-\log (\cos x)],dx}=\int_{{}}^{{}}{e^{x}[\tan x+\log (\sec x)]},dx $ $ =e^{x}\log (\sec x)+c $ $ { Since\int_{{}}^{{}}{e^{x}{ f(x)+{f}’(x) }dx=e^{x}f(x)+c} } $ .
BETA
