Some uncommon improper integral. Convergence. I bet the integral $\int_1^3\frac{dx}{\sqrt{\tan(x^3-7x^2+15x-9)}}$ converges, but have no idea about proof, except expansion of tan in a series near zeros of its argument (limits of the integration). 
 A: Factoring the polynomial in the function gives us
$$\sqrt{\tan(x^3-7x^2+15x-9)}=\sqrt{\tan((x-1)(x-3)^2)}$$
The tangent function can be bounded above as the following: $$\tan(y)=\frac{\sin(y)}{\cos(y)} < \frac{y}{\cos(y)} < 2y$$ when $y$ is sufficiently close to zero. Therefore  $$\sqrt{\tan((x-1)(x-3)^2)} < \sqrt{2(x-1)(x-3)^2} < \sqrt{6(x-3)^2} < \sqrt{6}|x-3|$$ for $x$ sufficiently close to 3. Therefore $$\frac{1}{\sqrt{\tan((x-1)(x-3)^2)}} > \frac{1}{\sqrt{6}|x-3|}$$
for $x$ sufficiently close to 3. This can you use to prove that the integral does not converge. 
A: If you used Taylor series, it is effectively clear since, close to $x=1$, $$\frac{1}{\sqrt{\tan(x^3-7x^2+15x-9)}}=\frac{1}{2 \sqrt{x-1}}+\frac{\sqrt{x-1}}{4}+O\left((x-1)^{3/2}\right)$$ does not make any problem while, close to $x=3$, $$\frac{1}{\sqrt{\tan(x^3-7x^2+15x-9)}}=\frac{1}{\sqrt{2} (x-3)}-\frac{1}{4 \sqrt{2}}+\frac{3 (x-3)}{32
   \sqrt{2}}+O\left((x-3)^2\right)$$ where the problem happens.
For sure, these are not the most pleasant expansions to perform. 
There is no doubt that wythagoras's approach is much simpler and effective.
