What is the arclength of the polar curve $f(theta) = thetasin(3theta)-sec^2theta$ over $theta in [0,pi/8]$ ?

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Answer 1 · 2018-06-01T14:07:34

$approx 0.41636$

we have
$r(x)=x*sin(3x)-sec^2(x)$
so
$(dr)/dx=sin(3x)+x*cos(3x)*3-2sec^2(x)*tan(x)$
I have only found a numerical value for this integral.

What is the arclength of the polar curve f(theta) = thetasin(3theta)-sec^2theta f(theta) = thetasin(3theta)-sec^2theta over theta in [0,pi/8] theta in [0,pi/8] ?