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Integration as Antiderivative

Given y= sin ...

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Given y= sin t^3 dy/dt=d[(sin t)^3]/dt dy/dt=3(sin t)^2 × d(sin t)/dt dy/dt=3(sin t)^2 × cos t explanation needed for step 2

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x = f^{''} (t) cos t + f^{'} (t) sin t

y = - f^{''} (t) sin t + f^{'} (t) cos t

\int {[{(\frac{d x}{d t})}^{2} + {(\frac{d y}{d t})}^{2}]}^{1 / 2} d t

x = f^{''} (t) cos t + f (t) sin t

y = - f^{''} (t) sin t + f^{'} (t) cos t .

\int {[{(\frac{d x}{d t})}^{2} + {(\frac{d y}{d t})}^{2}]}^{\frac{1}{2}} d t

x = a (cos t + t sin t)

y = a (sin t - t cos t), 0 < t < \frac{π}{2}

\frac{d^{2} x}{d t^{2}}, \frac{d^{2} y}{d t^{2}}

\frac{d^{2} y}{d x^{2}} .

\frac{d y}{d x}

\frac{d x}{d t} = a p cos p t, \frac{d y}{d t} = - b p sin p t

y = a cos t + b sin t

\frac{d^{2} y}{d t^{2}} =

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