Question

What is the definite integral here?

Answer 1

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

$\int\limits^{\pi/3}_{\pi/3} {4sec(t)(tan(t)-sec(t)} \, dx$=
$[\frac{4}{cos(t)} -4tan(t)]^{\pi/3}_{\pi/6}$=
$(\frac{4}{cos(\pi/3)} -4tan(\pi/3))-(\frac{4}{cos(\pi/6)} -4tan(\pi/6))$=
$4((\frac{1}{cos(\pi/3)} -1tan(\pi/3))-(\frac{1}{cos(\pi/6)} -1tan(\pi/6)))$=
$4(( \frac{1}{ \frac{1}{2} } - \sqrt{3} )-( \frac{1}{ \frac{ \sqrt{3} }{2}}- \frac{ \sqrt{3} }{3} ))$=
$4((2- \sqrt{3} )-( \frac{2}{ \sqrt{3} }- \frac{ \sqrt{3} }{3} ))$=
$\frac{-16 \sqrt{3}+24 }{3}$