Question #7ca15

1 Answer
Jan 12, 2018

tan^2 ( 2 x ) + 2 log | cos ( 2 x ) | + C

Explanation:

int 4 tan^3 ( 2 x ) dx =
= 4 int tan ( 2 x ) ( sec^2 ( 2 x ) - 1 ) dx =
= \underbrace{4 int tan ( 2 x ) sec^2 ( 2 x ) dx}_{I_1} \underbrace{- 4 int tan ( 2 x ) d x}_{I_2} .

I_1 = 4 int tan ( 2 x ) sec^2 ( 2 x ) dx = [ 2 int t dt ]_{ t = tan ( 2 x ) } =
= tan^2 ( 2 x ) + C

I_2 = - 4 int tan ( 2 x ) d x = -4 int sin ( 2 x ) / cos ( 2 x ) dx =
[ 2 int 1 / t dt ]_{t = cos ( 2 x )} = 2 log | cos ( 2 x ) | + C

NOTICE

As it turns out, I_1 has sec^2 ( 2 x ) as another solution, i.e.,
d / dx ( tan^2 ( 2 x ) ) = d / dx ( sec^2 ( 2 x ) ) = 4 tan ( 2 x ) sec^2 ( 2 x ) .