\(d/dx(F(X)) = d/dx(sin^2(\pi x/2))\) for \(0 \le x \le 1\)
We know that:
\(sin^2(\theta) = (1-cos(2\theta)) / 2\)
so:
\(d/dx(sin^2(\pi x/2)) = d/dx((1-cos(2 \pi x/2)) / 2) = d/dx(1/2 - cos(2(\pi x/2))/2) = d/dx\big(1/2 - cos(\pi x) / 2\big)\)
Therefore: \(f_X = \pi sin(\pi x)/2\)
For the probability that X < 1/4, we compute \(\int_{0}^{1/4} sin^2(\pi x/2) dx\)
Let’s intergrate:
\(1/2\int_{0}^{1/4} (1 - cos(\pi x)) dx = 1/2 \bigg[x - sin(\pi x)/ \pi \bigg]_{0}^{1/4} =\)
\(1/2\bigg(1/4 - sin(\pi/4)/ \pi\bigg) = 1/8 - \sqrt{2}/4 \pi\)