A curve in the $xy$-plane is given parametrically by $$x(t) = e^{2t}, \quad y(t) = e^{2t} \sin(2t), \quad t \in [0, \pi/2].$$ What is the length of this curve?
Ok, actually I know what to do, but I don't know how to do it because I can't get rid of the trigonometric term.
If it were $x(t) = e^{2t}\cos(t)$, I could have done it, but it's not so I can't get rid of the trigonometric term and integrate the expression.
With $x=e^{2t} \Rightarrow \dot{x} = 2 x$ and $y = e^{2t} \sin 2t = x \sin 2t \Rightarrow \dot{y} = \dot{x} \sin 2 t + 2 x \cos 2 t = 2 x \left(\sin 2 t + \cos 2 t\right)$, we want to calculate $$ \begin{eqnarray} \int_0^{\pi/2} dt \left({\dot{x}}^2 + {\dot{y}}^2\right)^{1/2} &=& \int_0^{\pi/2} dt \left[\left(2x\right)^2 + \left(2 x\right)^2 \left(\sin 2 t + \cos 2 t\right)^2\right]^{1/2} \\ &=& 2 \int_0^{\pi/2} dt \ e^{2t} \left[1 + \left(\sin 2 t + \cos 2 t\right)^2\right]^{1/2} \\ &=& \sqrt{2} \int_0^{\pi} du \ e^u \left(1 + \sin u \cos u\right)^{1/2} \end{eqnarray} $$ Wolfram cannot do that integral analytically.