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
Correct Answer: B
Solution :
Let \[\log (\sec x+\tan x)=t\Rightarrow \sec x\,dx=dt\] Therefore \[\int_{{}}^{{}}{\sec x\,\log (\sec x+\tan x)\,dx}=\int_{{}}^{{}}{t\,dt}\] \[=\frac{{{t}^{2}}}{2}+c=\frac{{{[\log (\sec x+\tan x)]}^{2}}}{2}+c.\]You need to login to perform this action.
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