<s>[INST] A problem of definite integral inequality? $$
\text { Minimum odd value of $a$ such that }\left.\left|\int_{10}^{19} \frac{\sin x d x}{\left(1+x^{a}\right)}\right|<\frac{1}{9} \text { is (where } a \in N\right)
$$
I proceed this way
As $|\sin x|<1$
Integration $|\frac{\sin x}{1+x^a}|<|\frac{1}{1+x^a}|$
And using trial and error method to get $a=3$
Any other way ??
 [/INST] Continuing from where you stopped we have that $$\left| \frac{1}{1+x^a}\right| \le \left| \frac{1}{1+10^a}\right|$$ on the interval given. Multiplying by $|\sin x|$ tells us that then integral is less than $$\int_{10}^{19}\frac{1}{1+10^a}\mathrm dx=\frac{9}{1+10^a}.$$
For this quantity to be less than $1/9$ we must have $${1+10^a}{9}>9,$$ or $1+10^a>81,$ which gives $$10^a>80.$$ The first integer $a$ satisfying this condition is $2.$ Hence the first odd such integer is $3.$
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