# "We want to find:" #
# \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad int_0^{pi/3} \ tan^5x tan^4x dx. #
# "We can proceed as follows:" #
# \qquad \quad int_0^{pi/3} \ tan^5x tan^4x \ dx \ = \ int_0^{pi/3} \ tan^9x \ dx #
# \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ \ = \ int_0^{pi/3} \ tan^8x tanx \ dx #
# \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ \ = \ int_0^{pi/3} \ ( tan^2x )^4 tanx \ dx #
# \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ \ = \ int_0^{pi/3} \ ( sec^2x - 1 )^4 tanx \ dx #.
# "Now applying the Binomial Theorem with power 4 to the" #
# "previous, we have, continuing:" #
# = \ int_0^{pi/3} \ ( [ sec^2x ]^4 + 3 [ sec^2x ]^3 ( - 1 )^1 + 6 [ sec^2x ]^2 ( - 1 )^2 #
# \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad + 3 [ sec^2x ]^1 ( - 1 )^3 + (-1)^4 )tanx \ dx #
# = \ int_0^{pi/3} \ ( sec^8x - 3 sec^6x + 6 sec^4x - 3 sec^2x + 1 )tanx \ dx #
# = \ int_0^{pi/3} \ ( [ sec^7x - 3 sec^5x + 6 sec^3x - 3 secx ] secx + 1 ) #
# \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \cdot tanx \ dx #
# = \ int_0^{pi/3} \ ( [ sec^7x - 3 sec^5x + 6 sec^3x - 3 secx ] secx tanx #
# \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \ + tanx ) \ dx #
# = \ int_0^{pi/3} \ [ ( sec^7x cdot ( secx tanx ) ) - 3 ( sec^5x cdot ( secx tanx ) ) #
# + 6 ( sec^3x cdot ( secx tanx ) ) - 3 ( secx cdot ( secx tanx ) ) ] + tanx ] \ dx #
# = \ int_0^{pi/3} \ [ sec^7x cdot ( secx tanx )- 3 sec^5x cdot ( secx tanx ) #
# + 6 sec^3x cdot ( secx tanx ) - 3 secx cdot ( secx tanx ) + tanx "]" \ dx #
# = [ 1/8 sec^8x - ( 3 cdot 1/6 sec^6x ) + ( 6 cdot 1/4 sec^4x ) #
# \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \ \ \ - ( 3 cdot 1/2 sec^2x ) - ln | cosx | ]_0^{pi/3} #
# = \ [ 1/8 sec^8x - 1/2 sec^6x + 3/2 sec^4x - 3 /2 sec^2x - \ ln | cosx | ]_0^{pi/3} #
\
# = \ [ 1/8 sec^8 ( pi/3 ) - 1/2 sec^6 ( pi/3 ) + 3/2 sec^4 ( pi/3 ) - 3 /2 sec^2( pi/3 ) #
# \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \quad - \ ln | cos ( pi/3 ) | ] #
# - [ 1/8 sec^8 ( 0 ) - 1/2 sec^6 ( 0 ) + 3/2 sec^4 ( 0 ) - 3 /2 sec^2( 0 ) - \ ln | cos ( 0 ) | ] #
# "Recall:" \qquad \qquad cos ( pi/3 ) = 1/2, \qquad sec pi/3 = 1 / { cos ( pi/3 ) } = 1 / { 1/2 } = 2; #
# \qquad \qquad \qquad \qquad \qquad \qquad \qquad cos ( 0 ) = 1, \quad sec( 0 ) = 1 / { 1 } = 1. #
# "Continuing:" #
# = \ [ 1/8 cdot 2^8 - 1/2 cdot 2^6 + 3/2 cdot 2^4 - 3 /2 cdot 2^2 -
ln | 1/2 | ] #
# \qquad \qquad \qquad \qquad \qquad \qquad \quad - [ 1/8 cdot 1^8 - 1/2 cdot 1^6 + 3/2 cdot 1^4 - 3 /2 cdot 1^2 -
\ ln | 1| ] #
# = \ [ color{red}cancel{2^5} - color{red}cancel{2^5} + ( 3 cdot 2^3 ) - ( 3 cdot 2^1 ) - [ - ln 2 ] ] #
# \qquad \qquad \qquad \qquad \qquad \qquad \quad - [ 1/8 - 1/2 + color{red}cancel{3/2} - color{red}cancel{3/2} - 0 ] #
# = \ [ 3 cdot 2 cdot ( 2^2 - 1 ) + ln 2 ] - 1/8 [ 1 - 4 ] #
# = \ [ 18 + ln 2 ] + 3/8 #
# = \ 57/8 + ln 2. #
# "Thus, at long last (!!): " #
# \qquad \qquad \qquad \qquad \qquad \quad int_0^{pi/3} \ tan^5x tan^4x \ dx \ = \ 57/8 + ln 2. \qquad \qquad \qquad \qquad \quad \ square #
