Find the smallest odd value of $k$ such that $\displaystyle\int_{10}^{19}{\sin x\over1+x^k}dx<\frac 19$ 
Find the smallest odd value of $k$ such that $\displaystyle\int_{10}^{19}{\sin x\over1+x^k}dx<\frac 19$

$$\displaystyle\begin{align*}\int_{10}^{19}{\sin x\over1+x^k}dx&<\int_{10}^{19}{\mid\sin x\mid\over1+x^k}dx\\&<\int_{10}^{19}{1\over1+x^k}dx\\&<\int_{10}^{19}{1\over1+10^k}dx\\&<{9\over1+10^k}\end{align*}$$
If $k=3$,
$$\displaystyle\begin{align*}\int_{10}^{19}{\sin x\over1+x^3}dx&<{9\over1+10^3}<\frac 19\end{align*}$$
If $k=1$,
$$\require{cancel}\displaystyle\begin{align*}\int_{10}^{19}{\sin x\over1+x}dx&<{9\over1+10}\bcancel{<}\frac 19\end{align*}$$
However this is not sufficient to prove that the inequality does not hold for $k=1$. Infact the inequality is actually true for $k=1$ (wolfram alpha). Can someone suggest a method to prove that the inequality is true for $k=1$. Thanks in advance.
EDIT-1
I have also tried,
$$\displaystyle\begin{align*}\int_{10}^{19}{\sin x\over1+x}dx&=\int_{10}^{4\pi}{\sin x\over1+x}dx+\int_{4\pi}^{5\pi}{\sin x\over1+x}dx+\int_{5\pi}^{6\pi}{\sin x\over1+x}dx+\int_{6\pi}^{19}{\sin x\over1+x}dx\\&<\int_{4\pi}^{5\pi}{\sin x\over1+x}dx+\int_{6\pi}^{19}{\sin x\over1+x}dx\\&<{5\pi-4\pi\over 1+10}+{19-6\pi\over 1+6\pi}\bcancel{<}\frac 19\end{align*}$$
Again I got stuck.
 A: You want to show that
$$\int_{10}^{19}\frac{\sin x dx}{1+x}\leq\frac19.$$
If this is true, it's because the "oscillations" of $\sin x$ cancel out, since the integral of the absolute value is a good bit larger than $1/9$. Since $1/(1+x)$ doesn't vary much in the interval $[10,19]$, you can subtract something close to its average, and see what happens if you then separate out the integrals. Use
$$\frac{1}{1+x}=\frac{1}{15}+\left(\frac{14-x}{15(1+x)}\right)$$
to get
$$\int_{10}^{19}\frac{\sin x dx}{1+x}=\frac1{15}\int_{10}^{19}\sin xdx+\int_{10}^{19}\frac{14-x}{15(1+x)}\sin xdx.$$
The first integral is $1/15$ of
$$-\int_{3\pi}^{10}\sin xdx+\int_{3\pi}^{6\pi}\sin xdx+\int_{6\pi}^{19}\sin xdx,$$
which is
$$-2+\int_0^{10-3\pi}\sin xdx+\int_0^{19-6\pi}\sin xdx.$$
Since $|\sin x|\leq x$, this is at most
$$-2+\frac{(10-3\pi)^2+(19-6\pi)^2}{2}<-2+\frac{0.58^2+0.16^2}{2}=-1.819.$$
The second integral is bounded by
$$\frac9{15}\max_{10\leq x\leq 19}\frac{|14-x|}{1+x}=\frac9{15}\frac4{11}=\frac{12}{55}.$$
So, the total integral is at most
$$-\frac{1.819}{15}+\frac{12}{55}<0.1<\frac19.$$
