The integral $∫e^x(tan\, x + log\, sec\, x)dx$ is equal to :
Answer & explanation
Correct answer: option 3
$I=∫e^x(\tan\, x + \log\, \sec\, x)dx$
$=∫e^x(f(x)+f'(x))dx$ form
here $f(x)= \log\sec x,f'(x)=\tan x$
so $I=e^x\log\sec x+C$
The integral $∫e^x(tan\, x + log\, sec\, x)dx$ is equal to :
Correct answer: option 3
$I=∫e^x(\tan\, x + \log\, \sec\, x)dx$
$=∫e^x(f(x)+f'(x))dx$ form
here $f(x)= \log\sec x,f'(x)=\tan x$
so $I=e^x\log\sec x+C$