Which of $\int_0^{0.5} \cos(x^2)\,dx$ and $\int_0^{0.5} \cos(\sqrt{x})\,dx$ is larger, and why? 
*

*$$\int_0^{0.5} \cos(x^2)\,dx$$

*$$\int_0^{0.5} \cos(\sqrt{x})\,dx$$
 A: If $0\le x\le 0.5$, $x^2\le\sqrt{x}\implies\cos x^2\ge \cos\sqrt{x}$, so the first integral is greater.
At DavidG's suggestion, I'll mention the $\implies$ uses the fact that $\cos x$ decreases on this interval.
A: Make a change of variables $x^2=u$ in the first integral (hence $dx=\frac{1}{2}u^{-1/2}du$) and $\sqrt x=u$ in the second integral (hence $dx=2udu$). Note that the integration limits change, but it is then easy to find the solution. Isn't?
A: The integral with $\cos x^2$ is slightly bigger than the one with $\cos \sqrt x$, simply for the property that x got a bigger exponent; you can verify this even on software like Wolfram Alpha Online.
edit: for the range considered this is true
A: Note that $\sqrt{0.5} < \pi/2$ and $0.5^2 < \pi/2$. So, $\cos(x^2)$ and $\cos(\sqrt{x})$ will be positive and decreasing for $x\in(0,0.5)$. Now note that $x^2$ is increasing less quickly than $\sqrt{x}$ on $x\in(0,0.5)$. In a sense this verifies $0 < \cos(\sqrt{x}) < \cos(x^2)$ for all $x\in(0,0.5)$. Hence, the integral of $\cos(x^2)$ will be greater than $\cos(\sqrt{x})$ over the interval $(0,0.5)$.
As an extra notion, it is not too difficult to extend this idea for the interval $(0,\sqrt{\pi/2})$. Going a bit further than this would require more work.
Hope this intuitive approach helps. If you have any questions feel free to ask them!
A: Just a small addition:
How do we know that for each $0\le x\le 0.5$ , $x^2\le\sqrt{x}$ ?
Two ways:


*

*Plot the graphs of the two functions on the same coordinate system. 
You will see immediately that the only point the graphs cross each other is $(x=0, y=0)$.


For each point other than $x=0$ you'll find out that $x^2\le\sqrt{x}$ .


*The easier way (or the "brutal" one...) :


Just assign the two edges of the range in the inequality equation:
Take a calculator.
Assign $x=0$ . You'll get an equality.
But if you assign higher values, including $x=0.5$ , you'll get through the values given by your calculator are higher for $\sqrt{x}$ than the values of $x^2$.
