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?
As the question is written, if you can show $\int _{10}^{19} \frac{\sin x}{1+x^3}dx \lt \frac 19$ you are done. But $\int _{10}^{19} \frac{\sin x}{1+x^3}dx \lt \int_{10}^{19}\frac 1{1001}dx=\frac {9}{1001}\lt \frac 19$ No need to evaluate the integral.