∫ e log ( tan x) dx is equal to :

Question

∫ e log ( tan x) dx is equal to : (A) log tan x + c (B) log sec x + c (C) tan x + c (D) e tan x + c

Tardigrade Admin · Accepted Answer

The given integral is I = ∫ e log( tan x)dx = ∫ tan x dx = ∫( sin x/ cos x) dx . cos x =t ⇒ - sin x dx = dt. So, I = -∫(dt/t) = - log t +c = log (1/t) + x = log (1/ cos x) + c = log sec x + c

Q. $\int e^{l o g (t a n x)} d x$ is equal to :

log tan x + c

log sec x + c

tan x + c

$e^{t a n x} + c$

The given integral is
$I = \int e^{l o g (t a n x)} d x = \int tan x d x$
$= \int \frac{s i n x}{c o s x} d x . cos x = t$
$\Rightarrow - sin x d x = d t .$
So, $I = - \int \frac{d t}{t} = - lo g t + c = lo g \frac{1}{t} + x$
$= lo g \frac{1}{c o s x} + c = lo g sec x + c$

Q. ∫elog(tanx)dx is equal to :