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Hopefully someone can help me understand the solution to the following integral which is the final step in a problem I'm looking at:

$\int^{\infty}_{0}\phi(x)\Phi(x)dx=\left[\frac{1}{2}\{\Phi(x)\}^2\right]^{\infty}_{0}=\frac{1}{2}\left[1^2-\left(\frac{1}{2}\right)^2\right]=\frac{3}{8}$

$\phi(x)$ is the probability density function of a Standard Normal and $\Phi(x)$ the distribution function. Obviously $\Phi(x)$ is the integral of $\int\phi(x)dx$ but I don't follow the first step in the evaluation of the above integral. The solution it's trivial. If anyone could explain the above it would be appreciated.

Thanks

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This is just standard integral substitution (sometimes called $u$-substitution). The fact that $\phi$ is the density and $\Phi$ is the cumulative distribution for a standard normal implies $$\Phi'(x) = \phi(x).$$ Therefore the integral can also be written as $$\int \Phi(x) \Phi'(x) \, dx.$$ And with the choice of substitution $$u = \Phi(x), \quad du = \Phi'(x) \, dx,$$ we see it becomes $$\int u \, du = \frac{u^2}{2} + C = \frac{(\Phi(x))^2}{2} + C.$$

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