$\int e^{x} [tan x - log (cos x)] d x =$

e^{x} log (sec x) + c

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e^{x} log (c o s e c x) + c

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e^{x} log (cos x) + c

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e^{x} log (sin x) + c

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Solution

The correct option is A $e^{x} log (sec x) + c$
$\int e^{x} (t a n x - l o g (c o s x)) d x = ?$
let $f (x) = - l o g (c o s x)$
$f^{'} (x) = - \frac{(- s i n x)}{c o s x}$
$f^{'} (x) = \frac{s i n x}{c o s x} = t a n x$
we know that $\int e^{x} (f (x) + f^{'} (x)) d x = e^{x} f (x) + c$
$\Rightarrow \int e^{x} (- l o g (c o s x) + t a n x) d x = e^{x} (- l o g (c o s x)) + c$
$= e^{x} l o g (s e c x) + c$

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