<s>[INST] Improper integral $\int_0^3 (1-x^{2}\sin(\frac{1}{x^{2}})) dx$ I have to determine the convergence of 
$\int_0^3 (1-x^{2}\sin(\frac{1}{x^{2}})) dx$
Can I say that
$(1-x^{2}\sin(\frac{1}{x^{2}}))\leq1+x^2$ 
and since $\int_0^3 1+x^2 dx$ 
is not even improper, the integral 
$\int_0^3 (1-x^{2}\sin(\frac{1}{x^{2}})) dx$ is convergent?
 [/INST] $$\int_0^3 \left(1-x^{2}\sin\left(\frac{1}{x^2}\right)\right)\ dx$$
$$=\int_0^3\ dx-\int_0^3 x^{2}\sin\left(\frac{1}{x^2}\right)\ dx$$
$$=3-\int_0^3 x^{2}\sin\left(\frac{1}{x^2}\right)\ dx$$
Since
$$\int_0^3 x^{2}\sin\left(\frac{1}{x^2}\right)\ dx \leq \int_0^3 x^2\ dx$$
And
$$\lim\limits_{t\to 0^+}\int_t^3 x^2\ dx=\frac13\lim\limits_{t\to 0^+} \left(3^3-t^3\right)=9$$
Therefore by the integral comparison test,
$$\int_0^3 x^{2}\sin\left(\frac{1}{x^2}\right)\ dx=\mbox{convergent}$$
Which implies that
$$\int_0^3 \left(1-x^{2}\sin\left(\frac{1}{x^2}\right)\right)\ dx=\mbox{convergent}$$
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