I am to determine the arc length of a function given in parametric form of the form
$$x = 50 (1 - \cos(t)) + 50 (2 - t) \sin(t)$$ $$y = 50 \sin(t) + 50 (2 - t) \sin(t)$$
I must determine the arc length of a function given in parametric form of the form in the interval $t=0$ to $t=2$ (it seems to me to be $2\pi$)<--this does not
$$\frac{dx}{dt}=50 (2 - t) \cos(t)$$
$$\frac{dy}{dt}=50 \cos(t) + 50 (2 - t) \cos(t) - 50 \sin(t)$$
$$\left (\frac{dx}{dt}\right )^{2}=2500((2-t)\cos(t))^2$$
$$\left (\frac{dy}{dt}\right )^{2}=2500 (\cos(t) + (2 - t) \cos(t) - \sin(t))^2$$
$$50\int_{0}^{2}\sqrt{((2-t)\cos(t))^2 + (\cos(t) + (2 - t) \cos(t) - \sin(t))^2 }dt$$
$$\:$$
(the result they propose is 100 units <--this does not), after a long time I cannot find an acceptable identity to solve this integral. Can you help me !!! Edit :50 was missing before the integral