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

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

1 Answer

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Answer 1 · 2018-06-01T14:07:34

$\approx 0.41636$ $approx 0.41636$

Explanation:

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

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

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What is the arclength of the polar curve f(θ)=θsin(3θ)−sec2θf(theta) = thetasin(3theta)-sec^2theta over θ∈[0,π8]theta in [0,pi/8] ?

1 Answer

Explanation:

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