Question 838142
Apparently you are integrating the function
{{{ 5sec(pi*x/6)=(30/pi)(pi/6)*sec(pi*x/6)}}}
to find the area under the curve between {{{x=0}}} and {{{x=2}}} .
step 1: {{{(30/pi)int((pi/6)*sec(pi*x/6),dx,0,2)=(30/pi)}}}{{{abs( "ln"abs(sec(pi*x/6)+tan(pi*x/6)))}}}{{{matrix(3,1,2," ",0)}}}
The antiderivative or indefinite integral of the function
{{{(pi/6)*sec(pi*x/6)}}} is the function
{{{f(x)="ln"abs(sec(pi*x/6)+tan(pi*x/6))}}}
The definite integral between the limits {{{x=0}}} and {{{x=2}}} is calculated as
{{{f(2)-f(0)}}} .
For {{{x=2}}} , {{{pi*x/6=pi*2/6=pi/red(3)}}} so {{{f(2)="ln"abs(sec(pi/red(3))+tan(pi/red(3)))}}}
For {{{x=0}}} , {{{pi*x/6=pi*0/6=0}}} so {{{f(0)="ln"abs(sec(0)+tan(0))="ln"abs(1+0)}}}
step 2: ={{{(30/pi)}}}{{{( "ln"abs(sec(pi/red(3))+tan(pi/red(3)))- "ln"abs(1+0))}}}
{{{sec(pi/3)=1/cos(pi/3)=1/(0.5)=highlight(2) }}} , {{{tan(pi/3)=highlight(sqrt(3)) }}} , and {{{"ln"abs(1+0))=ln(1)=0}}} , so (rounded to three decimal places). 
step 3: ={{{(30/pi)*ln(highlight(2)+highlight(sqrt(3)) )}}}
My calculator says that
step 4: ={{{highlight(12.576)}}}