Is the following integral convergent or divergent $\int_{0}^{1} \frac{1}{\sin x}dx$? Question: Is the following integral convergent or divergent $\int_{0}^{1} \frac{1}{\sin x}dx$? Use a comparative theorem to prove your results.
Answer attempt:
I want to know if my attempt at a solution is acceptable.
$$\int_{0}^{1} \frac{1}{\sin x}dx$$
By using a variable substitution:
$t = \sin x$,
$x = \arcsin t$,
$dx = \frac{1}{\sqrt{1-t^2}}dt$
we get: 
$$\int_{0}^{\sin 1} \frac{1}{t} \times \frac{1}{\sqrt{1-t^2}} dx$$
By using the following comparison:
$\frac{1}{t} \times \frac{1}{\sqrt{1-t^2}} \geq \frac{1}{t}$
This means that if the following integral is divergent we have proved that the original integral is also divergent:
$$\int_{0}^{\sin 1} \frac{1}{t}dx = \ln(\sin 1) - \ln(0)$$
$\ln(0)$ is undefined wich must mean that the original integral is divergent. 

Your attempt is fine. This is another approach, maybe simpler.
Note that $0<\sin(x)\leq x$ for $x\in (0,1]$, then for $0<t\leq 1$,
$$\int_t^1\frac{dx}{\sin(x)}\geq \int_t^1\frac{dx}{x}= -\ln(t)$$
which implies that 
$$\int_0^1\frac{dx}{\sin(x)}=\lim_{t\to 0^+}\int_t^1\frac{dx}{\sin(x)}\geq \lim_{t\to 0^+}(-\ln(t))=+\infty.$$
