<s>[INST] Maximum value of the integral: $\int _{10}^{19} \frac{\sin x}{1+x^a}dx$ 
Find the minimum odd value of $a$, where$a>1,~ a \in \mathbb{N}$ such that $$\int_{10}^{19}  \frac{\sin x}{1+x^a}dx<\frac{1}{9}$$

ATTEMPT:- Let $I(a)=\int _{10}^{19}  \frac{\sin x}{1+x^a}dx$ then from Leibnitz's rule, $$I'(a)=-\int _{10}^{19}  \frac{(\sin x)x^a\log x}{(1+x^a)^2}dx$$
Now using By Parts, $\int udv=uv-\int vdu$ with 
$$\begin{aligned}
u &= \sin x\log x &  &\implies &  du &= \cos x\log x+\frac{\sin x}{x} \\
dv &=\frac{x^a}{(1+x^a)^2} & &\implies &  v &=-\frac{1}{(1+x^a)\log a} \end{aligned} \\
\implies I'(a)=\sin x\log x\frac{1}{(1+x^a)\log a} -\int _{10}^{19} \frac{\cos x\log x+\frac{\sin x}{x}}{(1+x^a)\log a}$$
which doesn't seem solvable.
Next , since it asks for minimum odd value with respect to $a$, I put $a=3$ in the integral giving me:
$$I=\int _{10}^{19}  \frac{\sin x}{1+x^3}dx$$
The indefinite integral is in terms of SinIntegral and CosIntegral functions and even the definite integral is given as visual representation of the integral by Wolfram Mathematica 
The text says the answer is $3$.
How am I supposed to evaluate this integral?
 [/INST] Use the sequence of inequalities (taking into account $a>1$) $$\int _{10}^{19} \frac{\sin x}{1+x^a}dx <\int _{10}^{19} \frac{|\sin x|}{1+x^a}dx<\int _{10}^{19} \frac{1}{1+x^a}dx<\int _{10}^{19} \frac{1}{x^a}dx=\frac{19^{1-a}-10^{1-a}}{1-a}$$ Now, consider  $$f(a)= \frac{19^{1-a}-10^{1-a}}{1-a}$$ $$f(2)=\frac{9}{190}<\frac 19$$ $$f(3)=\frac{261}{72200}<\frac 19$$ $$f(4)=\frac{1953}{6859000}<\frac 19$$  So, the first odd integer value of $a$ is $3$.
Edit
However, and this makes me thinking that there could be some mistake in the question, if we perform a numerical integration $$g(a)=\int _{10}^{19} \frac{\sin x}{1+x^a}dx$$ the result is always negative $$g(2)=-0.0115882$$  $$g(3)=-0.0010402$$ $$g(4)=-0.0000963$$
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