Is $\int_{1}^{\infty}\frac{dx}{\sin^{5}x}$ finite or infinite? Is the definite integral given below finite or infinite? Why?
$$\int_{1}^{\infty}\frac{dx}{\sin^{5}x}$$
 A: The improper integral does not exist. Let 
$$I_B=\int_1^B \frac{1}{\sin^5 x}\,dx$$ 
The improper integral exists if $\lim_{B\to\infty} I_B$ exists. 
But $I_B$ itself fails to exist for arbitrarily large $B$, indeed already for $B=\pi$. We show that $\lim_{B\to \pi^-}I_B$ does not exist. Imagine calculating $I_{B-\epsilon}$ for small positive $\epsilon$. Make the change of variable $t=\pi-x$. Then
$$I_{B-\epsilon}=\int_{t=\epsilon}^{\pi-1} \frac{1}{\sin^5 t}\,dt.$$
Since $\sin t\le t$ in our interval, 
$$\int_{t=\epsilon}^{\pi-1} \frac{1}{\sin t^5}\,dt\ge  \int_{t=\epsilon}^{\pi-1} \frac{1}{t^5}\,dt.$$
But $\int_{t=\epsilon}^{\pi-1} \frac{1}{t^5}\,dt$ can be calculated explicitly, and blows up as $\epsilon$ approaches $0$ from the right.
One reason is enough, but there are other reasons that the integral does not exist. For example, let $f(x)=\frac{1}{\sqrt{\sin x}}$ when $\sin x$ is positive, and let $f(x)=- \frac{1}{\sqrt{|\sin x|}}$ when $\sin x$ is negative. Then there is no existence issue about $J_B=\int_1^B f(x)\,dx$. However, $J_B$ has period $2\pi$, so does not converge.  
