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