The problem I'm struggling with has asked me to find the $x$-representation of the half harmonic oscillator wave function with a potential of $\frac12kx^2$. Our setup started with the WKB approximation of the full harmonic oscillator, giving us this equation to start from: $$ \int_{0}^{x_0} \sqrt{2m}\sqrt{E_n-\frac12kx^2}dx=\left(n+\frac12\right)h $$ After some solving and using $\omega = \sqrt{\frac km}$, we eventually get to the normalized solution: $$\psi(x) = \langle x|n\rangle = 2^{-n/2}(n!)^{-1/2}\left(\frac{m\omega}{\hbar\pi}\right)^{1/4}H_n\left(\sqrt{\frac{m\omega}{\hbar}}\right)e^{-\frac{m\omega}{2\hbar}x^2}$$ where $H_n$ is the appropriate Hermite polynomial.
Now, I'm not expected to know what Hermite polynomials are at this level (first semester of 400-level Quantum Mechanics), but the way the professor wants me to answer this question is to see what I understand about the full harmonic oscillator and its solution above, and normalize it for the half harmonic oscillator. Essentially, we know $ \int_{-\infty}^{\infty}{\psi^*\psi} = 1$ and we want to show that $\int_{-\infty}^{\infty}\phi^*\phi$ is also equal to 1, where $\phi$ is the half harmonic oscillator wavefunction. In other words, we want to show $$\phi(x) = C^2\int_{-\infty}^{\infty}{\psi^*\psi} = 1$$ Intuitively, I know the answer is $\frac{1}{\sqrt{2}}$, and my professor confirmed this, but how am I supposed to get to this point? I only know the beginning and the end of the solution, but not the middle steps.