Past exam of the year 1 of the computer science course of the University of Oxford / 2015 / Discrete mathematics, continuous mathematics, and probability (resit) / 9 / a / ii / Solution

The cosine function is positive on $(0,\pi /2)$ and negative on $(\pi /2,2)$. Thus any nonzero $\alpha$ makes $\alpha \cos x$ negative on an interval of positive length. Taking $\alpha =0$ makes the total definite integral zero. HenceNormalization alone would suggest $\alpha =1/\sin 2$, but that candidate still violates nonnegativity and therefore is not a probability density function.