Proof: Clearly for $\displaystyle 1\le x\le\sqrt{\frac{\pi}{2}}$ it is obvious that $\displaystyle x\le x^2$. And since $\displaystyle \sin(x)$ is increasing on that interval it stands to reason that $\displaystyle \sin(x)\le\sin\left(x^2\right)\quad 1\le x\le\sqrt{\frac{\pi}{2}}$. Lastly noting that $\displaystyle \sin^2(x)\le\sin(x)\quad\forall x\in\mathbb{R}$. We may finally deduce that the proposed inequality is definitely true for $\displaystyle 1\le x\le\sqrt{\frac{\pi}{2}}$/ Therefore let us restrict our attention to $\displaystyle 0<x<1$. On this interval $\displaystyle x^2\le x$ and since $\displaystyle \cos(x)$ is decreasing on this interval we see that $\displaystyle \cos\left(x\right)\le\cos\left(x^2\right)\quad 0<x<1$. We need the following lemma.
Lemma: $\displaystyle \sin(x)<x\quad 0<x<1$
Proof: Let $\displaystyle \phi(x)=x-\sin(x)$. Then $\displaystyle \phi(0)=0$ and $\displaystyle \phi'(x)=1-\cos(x)$ and since $\displaystyle \cos(x)<1\quad 0<x<1$ the conclusion follows. $\displaystyle \blacksquare$
Using this lemma and the previous deductinos we see that $\displaystyle \cos(x)\le\cos\left(x^2\right)\implies \sin(x)\cos(x)\le\sin(x)\cos\left(x^2\right)\le x\cos\left(x^2\right)$. Now evaluating $\displaystyle \int_0^z \sin(x)\cos(x)dx\le\int_0^z x\cos\left(x^2\right)dx\quad 0<z<1$ gives us $\displaystyle \frac{1}{2}\sin^2(z)\le\frac{1}{2}\sin\left(z^2\ri ght)\quad 0<z<1$ and the result follows.