we can do
sec t=1/cos t
tan=(sint)/(cost)
then
4sect(tant-sect)=
4/cost(sint/cost-1/cost)=
4/cost((sint-1)/cost)=
[tex] \frac{4(sin(t)-1)}{cos^2(t)} [/tex]
[tex] \int\limits^{pi/3}_{pi/3} {\frac{4(sin(t)-1)}{cos^2(t)}} \, dt [/tex]
undistribute 4
[tex] 4 \int\limits^{pi/3}_{pi/3} {\frac{sin(t)-1}{cos^2(t)}} \, dt [/tex]
seperate
[tex] 4( \int\limits^{pi/3}_{pi/3} {\frac{sin(t)}{cos^2(t)}} \, dt-\int\limits^{pi/3}_{pi/3} {\frac{1}{cos^2(t)}} \, dt [/tex]
integrate seperately
[tex] \int\limits { \frac{sin(t)}{cos^2(t)} } \, dt [/tex]
u subsitute
u=cos(t)
du=-sin(t)
[tex] \int\limits { \frac{-1}{u^2} \, dt [/tex]
use the power rule since [tex] \frac{-1}{u^2}=-u^{-2} [/tex]
[tex]- \frac{u^{-1}}{-1} = \frac{1}{u} = \frac{1}{cos(t)} [/tex]
other part
[tex] \int\limits { \frac{1}{cos^2(t)} \, dt [/tex] =tan(t) as defined by the integral rule
now we got
[tex]4( \frac{1}{cos(t)} -tan(t))[/tex]=[tex]\frac{4}{cos(t)} -4tan(t))[/tex]
now we can do
[tex] \int\limits^{\pi/3}_{\pi/3} {4sec(t)(tan(t)-sec(t)} \, dx [/tex]=
[tex][\frac{4}{cos(t)} -4tan(t)]^{\pi/3}_{\pi/6}[/tex]=
[tex](\frac{4}{cos(\pi/3)} -4tan(\pi/3))-(\frac{4}{cos(\pi/6)} -4tan(\pi/6))[/tex]=
[tex]4((\frac{1}{cos(\pi/3)} -1tan(\pi/3))-(\frac{1}{cos(\pi/6)} -1tan(\pi/6)))[/tex]=
[tex]4(( \frac{1}{ \frac{1}{2} } - \sqrt{3} )-( \frac{1}{ \frac{ \sqrt{3} }{2}}- \frac{ \sqrt{3} }{3} ))[/tex]=
[tex]4((2- \sqrt{3} )-( \frac{2}{ \sqrt{3} }- \frac{ \sqrt{3} }{3} ))[/tex]=
[tex] \frac{-16 \sqrt{3}+24 }{3} [/tex]
