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?
$$\int_0^3 \left(1-x^{2}\sin\left(\frac{1}{x^2}\right)\right)\ dx$$ $$=\int_0^3\ dx-\int_0^3 x^{2}\sin\left(\frac{1}{x^2}\right)\ dx$$ $$=3-\int_0^3 x^{2}\sin\left(\frac{1}{x^2}\right)\ dx$$ 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 $$\int_0^3 \left(1-x^{2}\sin\left(\frac{1}{x^2}\right)\right)\ dx=\mbox{convergent}$$