Show that $\left| \frac{1}{2\pi}\int_s^t \frac{\sin Ku}{\sin u}du\right|\lt 1$ for $00$. Let $0<s<t\leq\frac{\pi}{2}$ and $K>0$. Show that 
$$\left| \frac{1}{2\pi}\int_s^t \frac{\sin Ku}{\sin u}du\right|\lt 1.$$
I got stuck on this problem using Bonnet's MVT or Taylor expansions. My approach was as follows:
Edit:
I thougt the case $K\leq1$ was as follows:
In this case, both $u,Ku\in(0,\frac{\pi}{2}]$ on which $\sin$ is increasing, so we have 
$$Ku\leq u\implies \sin(Ku)\leq\sin(u) \implies \frac{\sin(Ku)}{\sin(u)}\leq1.$$
Then, as $s,t\in(0,\frac{\pi}{2}]$, we have $s-t\lt \frac{\pi}{2}$ thus
$$\int_s^t \frac{\sin Ku}{\sin u}du\lt1*\frac{\pi}{2}<2\pi.$$
For $K>1$ I couldn't figure it out:
$\frac{1}{\sin(x)}$ is decreasing, and non-negative on $[0,\frac{\pi}{2}]$, so definitely on $[t,s]$. $\sin (Kx)$ is RI on $[t,s]$ so Bonnet's MVT shows that there is a $c\in(t,s)$ such that 
$$\int_s^t \frac{\sin Ku}{\sin u}du=\frac{1}{\sin(s)}\int_s^c \sin(Ku)du.$$
We can evaluate this integral, leading to
$$\int_s^t \frac{\sin Ku}{\sin u}du=\frac{1}{\sin(s)}*\frac{1}{K}\bigg(\cos(Ks)-\cos(Kc)\bigg)\leq\frac{2}{K\sin(s)}$$
as $\cos(Ks)$ is at most $1$, and $\cos(Kc)$ is at least $-1$.
Now I didn't know how to continue, as $\frac{1}{\sin(s)}$ goes to infinity as $s\to0$, so I can't find that this integral is bounded by $2\pi$. My second approach was looking at Taylor expansions but that just made the matter more complicated. Any help would be greatly appreciated!
 A: First, we may as well take $k> 1$ since you have already proved the conjecture for $0<k\leq 1$. Since the limit as $u$ approaches $0$
$$\lim_{u\to 0}\frac{\sin(ku)}{\sin(u)}=k$$
we know
$$\max_{u\in(0,\pi/2]}\left\{k,\frac{\sin(ku)}{\sin(u)}\right\}$$
exists and is well defined. Consider
$$u\in\left(0,\frac{\pi}{2k}\right]\subset \left(0,\frac{\pi}{2}\right)$$
Since $\frac{\pi}{2k}<\frac{\pi}{2}$ we know $\frac{\sin(ku)}{\sin(u)}$ is decreasing. This takes a second to prove, so start with the derivative:
$$\frac{d}{du}\left(\frac{\sin(ku)}{\sin(u)}\right)=\csc (u) (k \cos (k u)-\cot (u) \sin (k u))$$
If we assume this is zero for some $u$ in the interval, we get
$$0=\csc (u) (k \cos (k u)-\cot (u) \sin (k u))$$
Since $\csc(u)$ is positive for all $u\in(0,\pi/2)$, this implies
$$0=k \cos (k u)-\cot (u) \sin (k u)\Rightarrow k \cos (k u)=\cot (u) \sin (k u)$$
$$\Rightarrow 1=\frac{\cot (u) \tan (k u)}{k}$$
Differentiate both sides of this equation with respect to $k$ to get
$$0=\frac{\cot (u) \left(k u \sec ^2(k u)-\tan (k u)\right)}{k^2}$$
Again, $\cot(u)>0$ for $u\in (0,\pi/2)$. Therefore
$$0=k u \sec ^2(k u)-\tan (k u)$$
Set $y=ku$ (note that $y\in (0,\pi/2]$) to get
$$0=y \sec ^2(y)-\tan (y)$$
$$f(y)=\tan(y)\cos^2(y)-y=0$$
Note that $y=0$ solves this equation. Also, differentiating and setting equal to zero gives us
$$f'(y)=\cos^2(y)-\sin^2(y)-1=0$$
which implies the derivative changes signs at $y=2\pi n$ for $n\in\mathbb{Z}$. Since $f(1)=-0.545351<0$, $f(0)=0$, and $f'(y)\neq 0$ for $y\in (0,\pi/2]$, we can conclude $f(y)\neq 0$ for $y\in (0,\pi/2]$. As this is a contradiction, we can conclude
$$0\neq \csc (u) (k \cos (k u)-\cot (u) \sin (k u))$$
for $u\in \left(0,\frac{\pi}{2k}\right]$. Since $k>1$, for small $\epsilon>0$ this implies
$$\csc (u) (k \cos (k u)-\cot (u) \sin (k u))=\frac{1}{3} \left(k-k^3\right) \epsilon+O\left(\epsilon ^3\right)$$
$$=-\frac{1}{3} \left(k^3-k\right) \epsilon+O\left(\epsilon ^3\right)<0$$
We conclude $\frac{\sin(ku)}{\sin(u)}$ is decreasing for $u\in \left(0,\frac{\pi}{2k}\right]$. For 
$$u\in \left(\frac{\pi}{2k},\frac{\pi}{2}\right]$$
we use the fact that $\frac{2}{\pi}x\leq\sin(x)$ to get
$$\frac{\sin(ku)}{\sin(u)}\leq \frac{\pi}{2u}\leq k$$
Thus,
$$\max_{u\in(0,\pi/2]}\left\{k,\frac{\sin(ku)}{\sin(u)}\right\}=k$$
Now, define the zeros of the function $\sin(ku)$ as
$$a_n=\frac{n\pi}{k}$$
and define the integral
$$I_n=\int_{a_n}^{a_{n+1}}\frac{\sin(ku)}{\sin(u)}du$$
From the previous section of our proof, we have that
$$|I_n|=\left|\int_{a_n}^{a_{n+1}}\frac{\sin(ku)}{\sin(u)}du\right|\leq \int_{a_n}^{a_{n+1}}\left|\frac{\sin(ku)}{\sin(u)}\right|du \leq \int_{a_n}^{a_{n+1}}kdu=k(a_{n+1}-a_n)=\pi$$
Let us consider what happens when we add $I_n+I_{n+1}$ for odd $n$. Then
$$I_n+I_{n+1}=\int_{a_n}^{a_{n+1}}\frac{\sin(ku)}{\sin(u)}du+\int_{a_{n+1}}^{a_{n+2}}\frac{\sin(ku)}{\sin(u)}du$$
Note that $I_n$ is negative since $n$ odd corresponds to an odd zero of $\sin(ku)$ (implying that the function is decreasing). The same logic applies to say that $I_{n+1}$ is positive. Since $\sin(u)$ is increasing, we can $I_n$ from below with $a_{n}^{-1}$ and bound $I_{n+1}$ from above with $a_{n+2}^{-1}$. Thus
$$|I_n+I_{n+1}|=\left|\int_{a_n}^{a_{n+1}}\frac{\sin(ku)}{\sin(u)}du+\int_{a_{n+1}}^{a_{n+2}}\frac{\sin(ku)}{\sin(u)}du\right|$$
$$\leq \left|\frac{1}{a_n}\int_{a_n}^{a_{n+1}}\sin(ku)du+\frac{1}{a_{n+1}}\int_{a_{n+1}}^{a_{n+2}}\sin(ku)du\right|=\frac{4}{\pi -4 \pi  n^2}$$
Then we can bound any finite sum of these types of integrals by
$$\sum_{n\text{ odd}}^m|I_n+I_{n+1}|<-\sum_{n\text{ odd}}^\infty \frac{4}{\pi -4 \pi  n^2}=\frac{2}{\pi}<\frac{\pi}{2}$$
In much the same we, we can bound the evens by
$$\sum_{n\text{ even}}^m|I_n+I_{n+1}|<-2 + \frac{1}{\pi} + \pi<\frac{\pi}{2}$$
This one is slightly more difficult as $I_0$ and $I_1$ have to be counted separately due to division by zero otherwise. We now get to the finale of the proof. Obviously, there exists $i,j$ such that
$$a_i\leq s<a_{i+1}\text{ and }a_j<t\leq a_{j+1}$$
There are four cases: $i,j$ even, $i,j$ odd, $i$ even and $j$ odd, $i$ odd and $j$ even. We will present the first case as the rest follow in a similar manner. We have
$$\left|\frac{1}{2\pi}\int_{s}^{t}\frac{\sin(ku)}{\sin(u)}du\right|=\frac{1}{2\pi}\left|\int_{s}^{a_{i+1}}\frac{\sin(ku)}{\sin(u)}du+\sum_{m=i+1}^{j-2}\int_{a_m}^{a_{m+1}}\frac{\sin(ku)}{\sin(u)}du+\int_{a_{j-1}}^{t}\frac{\sin(ku)}{\sin(u)}du\right|$$
Now, note that
$$0<\int_{s}^{a_{i+1}}\frac{\sin(ku)}{\sin(u)}du\leq \pi$$
$$0>\int_{a_{j-1}}^{t}\frac{\sin(ku)}{\sin(u)}du>- \pi$$
This implies
$$-\pi\leq \int_{s}^{a_{i+1}}\frac{\sin(ku)}{\sin(u)}du+\int_{a_{j-1}}^{t}\frac{\sin(ku)}{\sin(u)}du\leq pi$$
Thus
$$\left|\frac{1}{2\pi}\int_{s}^{t}\frac{\sin(ku)}{\sin(u)}du\right|=\frac{1}{2\pi}\left|\int_{s}^{a_{i+1}}\frac{\sin(ku)}{\sin(u)}du+\sum_{m=i+1}^{j-2}\int_{a_m}^{a_{m+1}}\frac{\sin(ku)}{\sin(u)}du+\int_{a_{j-1}}^{t}\frac{\sin(ku)}{\sin(u)}du\right|$$
$$\leq \frac{1}{2\pi}\left| \pi \right|+\frac{1}{2\pi}\left| \frac{2}{\pi} \right|=\frac{1}{2}+\frac{1}{\pi ^2}=0.601321<1$$
From this, a proof strategy emerges: bound $s$ and $t$ between the zeros of $\sin(k u)$, bound the oscillating part between the zeros, show the ends are also bounded.
I would also like to add a clarification: there may be some bounds in this proof which are not as tight as possible, or I may have missed some details along the way. Unfortunately, these types of proofs are the most dull things to write, and I found my mind wandering for a significant portion of it. If anyone spots an error or a place that it could be tightened, please let me know. Of course, I am confident in the overall method, but have no wish to triple check all details. 
