Cho hàm số f ( x ) có f ( 0 ) = 0 và f ^\prime ( x ) = tan x + tan x $\forall$ x $\in$ ( 0; ) . Tính I = $\int$ f ( x ) dx
4 2

( 2) 0

1 1 - ln 2 1 1 - ln 2

f ^\prime ( x ) = tan 4 x + tan 2 x = tan 2 x (1 + tan 2 x ) .

tan 3 x
$\int$ f ^\prime ( x ) dx = $\int$ tan x (1 + tan x ) dx = $\int$ tan x d ( tan x ) = +C
2 2 2

3
tan 3 x
\implies f ( x) = +C .
3
tan 3 x
f ( 0) = 0 \implies C = 0 \implies f ( x ) =
3
$\pi$
$\pi$ 4 $\pi$
[ $\pi$ 4 $\pi$
[
3
1 | 1 |
dx = $\int$ tan x ( tan x + 1) dx - $\int$ tan xdx | = $\int$ tan xd ( tan x ) - $\int$ tan xdx |
4 4 4
tan x
I = $\int$ 2

3 3 0 3 0
0
0
|] 0
|]

1 tan 2 x $\pi$ 4 $\pi$
[ 1 1 2 [ 1 - ln 2
= |0 + ln cos x |04 | = + ln |= .
3 2 ] 3 2 2 ] 6