Question 5 is a hard question, so we will deal with it in parts. In the fourth part we computes the expectation of momentum squared, it is similar to the previous calculation by not quite.
$ \begin{eqnarray*} \langle p^2 \rangle &=& \int\limits_{0}^{L}{\psi(x, t)^* i\hbar\frac{\partial}{\partial x}(i\hbar\frac{\partial}{\partial x} \psi(x, t)) dx} \\ &=& -\hbar^2\int\limits_{0}^{L}{\psi(x, t)^* \frac{\partial^2}{\partial x^2} \psi(x, t)) dx} \\ &=& -\hbar^2\int\limits_{0}^{L}{(\sqrt{\frac{2}{L}}\sin(\frac{n\pi x}{L})e^{-\frac{i E_n t}{\hbar}})^* \frac{\partial^2}{\partial x^2} (\sqrt{\frac{2}{L}}\sin(\frac{n\pi x}{L})e^{-\frac{i E_n t}{\hbar}}) dx} \\ &=& -\hbar^2\int\limits_{0}^{L}{(\sqrt{\frac{2}{L}}\sin(\frac{n\pi x}{L})e^{\frac{i E_n t}{\hbar}}) \frac{\partial^2}{\partial x^2} (\sqrt{\frac{2}{L}}\sin(\frac{n\pi x}{L})e^{-\frac{i E_n t}{\hbar}}) dx} \\ &=& -\hbar^2\int\limits_{0}^{L}{(\sqrt{\frac{2}{L}}\sin(\frac{n\pi x}{L})) \frac{\partial^2}{\partial x^2} (\sqrt{\frac{2}{L}}\sin(\frac{n\pi x}{L})) dx} \\ &=& -\hbar^2\int\limits_{0}^{L}{(\sqrt{\frac{2}{L}}\sin(\frac{n\pi x}{L})) (-(\frac{n\pi}{L})^2) (\sqrt{\frac{2}{L}}\sin(\frac{n\pi x}{L})) dx} \\ &=& \frac{2n^2\hbar^2\pi^2}{L^3}\int\limits_{0}^{L}{\sin^2(\frac{n\pi x}{L}) dx} \\ \end{eqnarray*} $
We let $ y = \frac{2 n \pi x}{L} $, when $ x = 0 $, we have $ y = 0 $, when $ x = L $, we have $ y = 2 n \pi $, lastly $ dy = \frac{2 n \pi}{L} dx $.
$ \begin{eqnarray*} \langle p^2 \rangle &=& \frac{2n^2\hbar^2\pi^2}{L^3}\int\limits_{0}^{L}{\sin^2(\frac{n\pi x}{L}) dx} \\ &=& \frac{2n^2\hbar^2\pi^2}{L^3}\int\limits_{0}^{2n\pi}{\sin^2(y) \frac{L}{2n\pi}dy} \\ &=& \frac{n\hbar^2\pi}{L^2}\int\limits_{0}^{2n\pi}{\sin^2(y) dy} \\ &=& \frac{n\hbar^2\pi}{L^2}\int\limits_{0}^{2n\pi}{\frac{1 - \cos(2y)}{2} dy} \\ &=& \frac{n\hbar^2\pi}{2L^2}\int\limits_{0}^{2n\pi}{(1 - \cos(2y)) dy} \\ &=& \frac{n\hbar^2\pi}{2L^2}(y - \sin(2y)|_{0}^{2n\pi} \\ &=& \frac{n^2\hbar^2\pi^2}{L^2} \end{eqnarray*} $
Therefore the expectation of the momentum squared operator is $ \frac{n^2\hbar^2\pi^2}{L^2} $
$ \begin{eqnarray*} \langle p^2 \rangle &=& \int\limits_{0}^{L}{\psi(x, t)^* i\hbar\frac{\partial}{\partial x}(i\hbar\frac{\partial}{\partial x} \psi(x, t)) dx} \\ &=& -\hbar^2\int\limits_{0}^{L}{\psi(x, t)^* \frac{\partial^2}{\partial x^2} \psi(x, t)) dx} \\ &=& -\hbar^2\int\limits_{0}^{L}{(\sqrt{\frac{2}{L}}\sin(\frac{n\pi x}{L})e^{-\frac{i E_n t}{\hbar}})^* \frac{\partial^2}{\partial x^2} (\sqrt{\frac{2}{L}}\sin(\frac{n\pi x}{L})e^{-\frac{i E_n t}{\hbar}}) dx} \\ &=& -\hbar^2\int\limits_{0}^{L}{(\sqrt{\frac{2}{L}}\sin(\frac{n\pi x}{L})e^{\frac{i E_n t}{\hbar}}) \frac{\partial^2}{\partial x^2} (\sqrt{\frac{2}{L}}\sin(\frac{n\pi x}{L})e^{-\frac{i E_n t}{\hbar}}) dx} \\ &=& -\hbar^2\int\limits_{0}^{L}{(\sqrt{\frac{2}{L}}\sin(\frac{n\pi x}{L})) \frac{\partial^2}{\partial x^2} (\sqrt{\frac{2}{L}}\sin(\frac{n\pi x}{L})) dx} \\ &=& -\hbar^2\int\limits_{0}^{L}{(\sqrt{\frac{2}{L}}\sin(\frac{n\pi x}{L})) (-(\frac{n\pi}{L})^2) (\sqrt{\frac{2}{L}}\sin(\frac{n\pi x}{L})) dx} \\ &=& \frac{2n^2\hbar^2\pi^2}{L^3}\int\limits_{0}^{L}{\sin^2(\frac{n\pi x}{L}) dx} \\ \end{eqnarray*} $
We let $ y = \frac{2 n \pi x}{L} $, when $ x = 0 $, we have $ y = 0 $, when $ x = L $, we have $ y = 2 n \pi $, lastly $ dy = \frac{2 n \pi}{L} dx $.
$ \begin{eqnarray*} \langle p^2 \rangle &=& \frac{2n^2\hbar^2\pi^2}{L^3}\int\limits_{0}^{L}{\sin^2(\frac{n\pi x}{L}) dx} \\ &=& \frac{2n^2\hbar^2\pi^2}{L^3}\int\limits_{0}^{2n\pi}{\sin^2(y) \frac{L}{2n\pi}dy} \\ &=& \frac{n\hbar^2\pi}{L^2}\int\limits_{0}^{2n\pi}{\sin^2(y) dy} \\ &=& \frac{n\hbar^2\pi}{L^2}\int\limits_{0}^{2n\pi}{\frac{1 - \cos(2y)}{2} dy} \\ &=& \frac{n\hbar^2\pi}{2L^2}\int\limits_{0}^{2n\pi}{(1 - \cos(2y)) dy} \\ &=& \frac{n\hbar^2\pi}{2L^2}(y - \sin(2y)|_{0}^{2n\pi} \\ &=& \frac{n^2\hbar^2\pi^2}{L^2} \end{eqnarray*} $
Therefore the expectation of the momentum squared operator is $ \frac{n^2\hbar^2\pi^2}{L^2} $
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