<s>[INST] 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). 
 [/INST] 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. 
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