How many ways can you show that $ \int_0^1 \frac{\sin(\pi x)}{(1-x)^2}\,\mathrm{d}x$ is divergent? How many ways can you show that this integral is divergent? $\displaystyle\int_0^1 \dfrac{\sin(\pi x)}{(1-x)^2} \,dx$
The only way I was able to show this was using a hint which was given to me that I don't think I could've come up with. That hint was to observe that $\large \frac{\sin(\pi x)}{1-x}$ is bounded on $[0,1]$ and then use substitution followed by the $p$-test for $\displaystyle \int_0^1 \frac{M}{1-x}$, where $M$ is the upper bound.  
Any tips or new skills would be appreciated!  
Someone mentioned comparing $\sin(\pi x)$ to $\pi x (1-x)$ though I can't seems to figure this out either (even with the help of the series expansion).
 A: Recalling the Taylor series of $\sin(\pi x)$ at the point $x=1$

$$ \sin(\pi x) = -\pi \, \left( x-1 \right) + \frac{\pi^3}{3!}\left( x-1 \right) ^{3}-\dots \implies \sin(\pi x)\sim  -\pi \, \left( x-1 \right)$$

which implies the integrand behaves like 

$$ \frac{\sin(\pi x)}{(x-1)^2} \sim \frac{-\pi(x-1)}{(x-1)^2}= \frac{-\pi}{x-1} $$

Now, I think you can finish the problem. See my answer.
A: The problem is a little artificial the way it is formulated.  It is more convenient to make a change of variables $t=1-x$ and then you get a similar integral with a singularity at the origin.  Your integral converges if and only if the new one does, therefore it is sufficient to show that $\int_0^1 \frac{\sin t}{t^2}$ diverges.  This follows by comparing $\sin t$ to $t$.
A: Do the substitution $x\mapsto 1-x$; use periodicity of $\sin$; do a Taylor expansion of $\sin$; observe that $\int_0 \frac{1}{x}\mathrm{d}x$ is divergent.
A: Since you have one proof of the theorem, it follows that there is a countably infinite number of other proofs. I believe this still holds if you exclude proofs that have useless deductions (to eliminate redundancy).
