Question

Question #549f2

Answer 1

`e^(tan x) sinx +C`

Explanation:

`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.)