Determining if these integrals converges or diverges. C.T. is comparison test
TYPE II is when a improper integral is improper but not at $\infty$. 
a)
$$\int_{1}^{\infty} \frac{\sin\left(\frac{\pi}{x}\right)}{x^2}dx$$
Let g(x) = $\frac{1}{x^2}$ because $|sin(\frac{\pi}{x})| \leq 1$. Since the numerator has been maximized and denominator has been minimized and we also know that g(x) is clearly always positive, then it fits the CT Definition of $$0 \leq f(x) \leq g(x)$$
$$\int_{1}^{\infty} \frac{1}{x^2} dx$$ Always converges.
Since $\int_{1}^{\infty} g(x)dx$ converges, then $\int_{1}^{\infty} f(x)dx$ converges by C.T
b) $$\int_{0}^{2} \frac{\sec^2 (x)}{x\sqrt{x}}dx$$
Yeah, uhm I have no idea. I'm pretty sure this has many discontinuities because this can be rewritten too 
$f(x) = \frac{1}{\cos^2 (x) x \sqrt{x}}$ which is discontinues at x = 0 and x = $\frac{\pi}{2}$. I don't think that fits C.T., I don't know how to integrate this so I don't know how I would use the TYPE II definition on this. 
c) $\int_{0}^{\infty} \frac{x}{x^3+1} dx$
Could I let $g(x) = \frac{x}{x^3} = \frac{1}{x^2}$  and say the same conclusion as a)? 
Sorry my textbook has little to none questions on C.T. and I don't get these
 A: For the second problem, it is correct to say that $f(x)=\frac{\sec^2(x)}{x^{3/2}}$ has discontinuities at $x=0$ and $x=\pi/2$.  But a pair of discontinuities alone does not render a function non-integrable.  However, these discontinuities are also singularities of $f$ at $x=0$ and $x=\pi/2$.  
For the singularity at $x=0$, we see that 
$$f(x)\ge \frac{1}{x^{3/2}}$$
and $\int_0^2\frac{1}{x^{3/2}}\,dx$ diverges.  That is enough to show that the integral diverges.
Aside, we see that $\int_{1}^2f(x)\,dx$ also diverges since the term $\sec^2(x)= \frac{1}{(x-\pi/2)^2}+O(1)$ as $x\to \pi/2$ and $$\begin{align}\int_1^2\frac{\sec^2(x)}{x^{3/2}}\,dx&\ge \frac{1}{2^{3/2}}\int_1^2 \sec^2(x)\,dx\\\\&=\frac{1}{2^{3/2}}\int_1^2\left(\frac{1}{(x-\pi/2)^2}+O(1)\right)\,dx \end{align}$$
where the last integral clearly diverges by comparison.
A: a) Since $\left|\sin(\cdot)\right|\leq 1$ the given integral is absolutely convergent and by the substitution $x=\frac{1}{z}$:
$$ \int_{1}^{+\infty}\frac{\sin\left(\frac{\pi}{x}\right)}{x^2}\,dx = \int_{0}^{1}\sin(\pi z)\,dz = \color{red}{\frac{2}{\pi}},$$
sic et simpliciter;
b) The function $\frac{1}{x\sqrt{x}\cos^2(x)}$ has a non-integrable singularity (branch point of the $x^{-3/2}$ kind) in a right neighbourhood of the origin and a double pole at $x=\frac{\pi}{2}$, so it is not integrable over $(0,2)$ by two different reasons;
c) The function $\frac{x}{x^3+1}$ is integrable over $\mathbb{R}^+$ (since it is positive and $\leq\min(x,x^{-2})$) and
$$ \int_{0}^{+\infty}\frac{x\,dx}{x^3+1}=\color{red}{\frac{2\pi}{3\sqrt{3}}} $$
can be shown in a good number of ways.
