which function and integral are greater problem My question is about kind of comparing values in a function, for example in this question
given 'c' a positive constant we have the following function $f(x)=cx^3+x^5-1$ , determine which is greater $f(\sin(x))$ or $f(\cos(x))$ for the values $\frac{\pi}{4}$$<x<$$\frac{\pi}{3}$
my try was that knowing how the graphs look, we know that in $\frac{\pi}{4}$$<x<$$\frac{\pi}{3}$
the slope of $\sin(x)$ is positive while the slope of $\cos(x)$ is negative and they both intersect at $\frac{\pi}{4}$ so my guess was that  $\sin(x)$ would be greater(would appreciate if anyone can confirm or have another way).
the second part of the question was similar we also need to determine if the integrals are equal or find which is smaller or greater.
$$\int_1^{\frac{\pi}{2}}\cos^8(x) dx $$ and $$\int_\frac{\pi}{2}^{\pi}\cos^8(x) dx $$
how can I find out which of these is greater/equal or less than the other? for the first part, I think it was easier because there was a way to compare since it is a $\sin(x)$ and $\cos(x)$ but I seem to be lost with the integrals.
So I am asking for advice and tips on how to approach and solve these questions, the answer is not important as I have a lot of these questions I just want to practice the topic and find a way to approach it.
 A: $f(x) = x^5 + cx^3 - 1$
$f'(x) = 5x^4 + 3cx^2$
We can see that $f'(x) \geq 0 \forall x \in \Bbb R$, therefore $f(x)$ is increasing for all x.
From the graphs of $\cos x$ and $\sin x$, one can clearly mention that they both intersect at $x=\pi/4$, after which for $x \in (\pi/4, \pi/3)$
$\sin x > \cos x$
$\implies f(\sin x) > f(\cos x)$

We know that the graph of $\cos x$ looks like this for $x \in (0,\pi)$:

So, we can say that graph of any even power of $\cos x$ would look something like this:

Now, because $\cos^8(\frac{\pi}{2} + x) = cos^8(\frac{\pi}{2} - x)$, We can say that $\cos^8x$ is symmetrical about $x=\pi/2$
Therefore, $\int_0^{\pi/2}\cos^8xdx = \int_{\pi/2}^{\pi}\cos^8xdx$
Now, As $\cos^8x$ is always non-negative, Area bounded by it in $x \in (0,\pi/2)$ would be greater than the area bounded by it in $x \in (1,\pi/2)$
$$\implies \int_0^{\frac{\pi}{2}}\cos^8xdx > \int_1^{\frac{\pi}{2}}\cos^8 xdx$$
$$\implies \int_{\frac{\pi}{2}}^{\pi}\cos^8xdx > \int_1^{\frac{\pi}{2}}\cos^8 xdx$$
A: Hints

*

*For the first problem, note that for $\ x\ge0\ $, $\ f(x)\ $ is a strictly increasing function of $\ x\ $, so the larger of $\ f(\sin\theta)\ $ or $\ f(\cos\theta)\ $ is completely determined by which is the larger of $\ \sin\theta\ $ and $\ \cos\theta\ $.  Over the interval $\ \frac{\pi}{4}\le\theta\le\frac{\pi}{3}\ $ $\ \sin\theta\ $ strictly increases from $\ \frac{1}{\sqrt{2}}\ $ to $\ \frac{\sqrt{3}}{2}\ $ while $\ \cos\theta\ $ strictly decreases from $\ \frac{1}{\sqrt{2}}\ $ to $\ \frac{1}{2}\ $. Therefore
$$
\frac{1}{2}<\cos{\theta}<\frac{1}{\sqrt{2}}<\sin\theta<\frac{\sqrt{3}}{2}
 $$ for all $\ \theta\in\left(\frac{\pi}{4},\frac{\pi}{3}\right)\ $.

*For the second problem, note that $\ \cos(\pi-\theta)=-\cos\theta\ $ and therefore $\ \cos^8(\pi-\theta)=\cos^8\theta\ $. Therefore, a change of variables in the integral $\ \int_\limits{\frac{\pi}{2}}^\pi\cos^8(x)dx\ $ from $\ x\ $ to $\ y=\pi-x\ $ gives
$$
  \int_\limits{\frac{\pi}{2}}^\pi\cos^8(x)dx=\int_0^\frac{\pi}{2}\cos^8(y)dy\ .
 $$
