Question #7ca15

1 Answer
paho
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 ) .#

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