Integral Calculus Question 147
Question: $ \int_{{}}^{{}}{\sec x\log (\sec x+\tan x)\ dx=} $
Options:
A) $ {{[\log (\sec x+\tan x)]}^{2}}+c $
B) $ \frac{1}{2}{{[\log (\sec x+\tan x)]}^{2}}+c $
C) $ {{\sec }^{2}}x+\tan x\sec x+c $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
Let $ \log (\sec x+\tan x)=t\Rightarrow \sec x,dx=dt $ Therefore $ \int_{{}}^{{}}{\sec x,\log (\sec x+\tan x),dx}=\int_{{}}^{{}}{t,dt} $ $ =\frac{t^{2}}{2}+c=\frac{{{[\log (\sec x+\tan x)]}^{2}}}{2}+c. $
BETA
