$$ \text { Minimum odd value of $a$ such that }\left.\left|\int_{10}^{19} \frac{\sin x d x}{\left(1+x^{a}\right)}\right|<\frac{1}{9} \text { is (where } a \in N\right) $$
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 ??
On
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 $3.$
$$\left|\int_{10}^{19} \frac{\sin x \,dx}{1+x^{a}}\right|\le \int_{10}^{19}\left|\frac{\sin x }{1+x^{a}}\right|\,dx\le \int_{10}^{19}\frac{1 }{1+x^{a}}\,dx<\int_{10}^{19}\frac{1}{x^a}\,dx= \frac{19^{1-a}-10^{1-a}}{1-a}.$$
Since the denominator contains $1-a$, we will assume that $a >1$. For $a=2$,
$$ \frac{19^{-1}-10^{-1}}{-1}=\frac{9}{190}<\frac19.$$
For $a =3$,
$$\frac{19^{-2}-10^{-2}}{-2}=\frac{261}{72200}<\frac19.$$
Therefore, the smallest odd value of $a$ which satisfies the inequality is $a=3$.