Question #549f2

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Answer 1 · 2016-09-29T15:55:13

$e^(tan x) sinx +C$

$int e^(tan x) (secx - sinx) dx = int e^(tanx)sec x dx - int e^(tan x) sin x dx$

For the first integral, rewrite and use integration by parts:

$int e^(tanx)sec x dx = int underbrace(cos x)_u underbrace(e^(tan x )sec^2 x dx)_(dv)$

$= cos x e^(tanx) - int e^(tanx )(-sin x ) dx$

$= cos x e^(tanx) + int e^(tanx ) sin x dx$

So now we have

$int e^(tan x) (secx - sinx) dx = [int e^(tanx)sec x dx] - int e^(tan x) sin x dx$

$= [cos x e^(tanx) + int e^(tanx )(sin x ) dx] - int e^(tan x) sin x dx$

When we simplify, we get

$int e^(tan x) (secx - sinx) dx = cos x e^(tanx) +C$

(Check the answer by differentiating.)