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Given that $\Phi(x)=\int_{-\infty}^{x}\phi(u)\, du$ and that $\phi(u) = \frac{1}{\sqrt{2\pi}}e^{-u^2/2}$.

Using the fact that $\lim_{x\rightarrow\infty}\Phi(x)=1$ and using the properties of the integral as well as a substitution show that $\Phi(-x)=1-\Phi(x)$ (We can assume that $x>0$)

I've attempted taking the integral, to no avail, as $e^{-u^2/2}$ is not integrable (at least not to my knowledge) and I've thought about using the property $\frac{d}{dx}\left[\int_{0}^{b(x)}f(t)\, dt\right] = f(b(x))b'(x)$, but i'm not entirely sure if it'd be correct to approach it that way.

I.e. doing something like this: $\frac{d}{dx}\left[\int_{-\infty}^{0}\phi(u)\, du + \int_{0}^{x}\phi(u)\, du\right]$

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    Did you mean $\phi(u)=\frac{1}{\sqrt{2\pi}}\,\exp\left(-\frac{u^2}{2}\right)$? – 2017-01-19
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    I did indeed, thank you. Edited. – 2017-01-19
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    Please provide your thoughts and attempt, otherwise your question will be closed. – 2017-01-19
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    Just done so. Thank you – 2017-01-19

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Note that $\phi$ is a symmetric function, i.e $$ \phi(u)=\phi(-u), \quad u \in \Bbb R. $$ Using this and the fact that $\lim_{x\to \infty}\Phi(x)=1$ we get that \begin{align} \Phi(-x) &= \int_{-\infty}^{-x} \phi(u) \;du = \int_{-\infty}^{\infty}\phi(u) \;du - \int_{-x}^\infty \phi(u) \;du = 1-\int_{-x}^\infty \phi(u) \;du \\[6mm] &= 1 + \int_{x}^{-\infty} \phi(-u) \;du =1- \int_{-\infty}^{x} \phi(-u) \;du = 1-\int_{-\infty}^{x} \phi(u) \;du = 1- \Phi(x). \end{align}

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    I follow now, thank you. – 2017-01-19