Question

Question #7ca15

Answer 1

`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 ) .`