I was working on a practice exam today and I got stumped on the following question:
The continuous random variable $Y$ has probability density function $f$ given by $$f(y)=\cos(y),\quad 0\le y\le\pi/2$$ $$f(y)=0,\quad \text{elsewhere}$$ The random variable $U$ is defined by $U=Y^n,\ n\in\mathbb{Z}^+$. Find the CDF of $U$ valid for $0\le u\le\left(\frac{\pi}{2}\right)^n$.
At first I thought, $P(U=u)=P(Y^n=u)=P(Y=u^{1/n})=\cos(u^{1/n})$ so the CDF is $$H_1(u)=\int_0^u\cos(t^{1/n})dt$$ However, this is a pain to integrate and so I decided to try a different approach. Since $F(y)=\sin(y)$ we know that $P(U\le u)=P(Y\le u^{1/n})=\sin(u^{1/n})$ and so the CDF of $U$ is $$H_2(u)=\sin(u^{1/n})$$ While taking the test I was inclined to believe that the latter solution ($H_2(u)$) was correct as it satisfied the bounds stated in the question. Indeed after checking the mark scheme, $H_2(u)$ is the given answer. However, I fail to understand why the former approach is incorrect. Could somebody explain why $H_2(y)$ is right but $H_1(y)$ is wrong?