Convergence of $\int^a_0\frac{1}{(\sin(t-1))^3}dt$

Question

I'm trying to study the convergence of $$\int^a_0\frac{1}{(\sin(t-1))^3}dt$$ In particular I'm trying to find the value of $a$ that permits the convergence. I know that $1$ is the critic point but I don't know if $a$ have to be $a>1$ or $a<1$ or even $a=1$. I don't know the consequences of the three conditions on convergence of this improper integral. Can someone explain me clearly the solution and the consequences that each three conditions have on the integral? Thank you in advance!!! I prefer a method that does not require computing the integral itself

Answer 1

We have

$$\frac1{(\sin(t-1))^3}\underset{t\to1}\sim \frac1{(t-1)^3}$$ and clearly the integral

$$\int_0^1\frac{dt}{(t-1)^3}$$ is divergent so for $a\ge1$ the given integral is divergent. For $0\le a<1$ the integral is convergent since the function is continuous on the interval $[0,a]$.