<s>[INST] $\displaystyle\int\limits_{\frac{\pi}{2}}^{\pi}\dfrac{1}{(\sin\theta - 2\cos\theta)^2}\,d\theta$: singularity in Weierstrass sub Could I use the integration by Weierstrass substitution or tangent-half angle substitution for this case? Because I see on Wikipedia, when the lower limit was $0$, they made the definite integral split into two things, where the first integral is from $0$ to $\pi$ and the second one is $\pi$ to $2 \pi$. Because the article said that "In the first line, one cannot simply substitute $t=0$ for both limits of integration."
This is what the article work like this

Blockquote
$$\begin{align}
\int_0^{2\pi}\frac{dx}{2+\cos x}
&= \int_0^\pi \frac{dx}{2+\cos x} + \int_\pi^{2\pi} \frac{dx}{2+\cos x} \\[6pt]
&=\int_0^\infty \frac{2\,dt}{3 + t^2} + \int_{-\infty}^0 \frac{2\,dt}{3 + t^2} & t &= \tan\tfrac x2 \\[6pt]
&=\int_{-\infty}^\infty \frac{2\,dt}{3+t^2} \\[6pt]
&=\frac{2}{\sqrt 3}\int_{-\infty}^\infty \frac{du}{1+u^2} & t &= u\sqrt 3 \\[6pt]
&=\frac{2\pi}{\sqrt 3}.
\end{align}$$
In the first line, one cannot simply substitute
${\textstyle t=0}$
for both limits of integration. The singularity (in this case, a vertical asymptote) of
${\textstyle t=\tan {\tfrac {x}{2}}}$
at
${\textstyle x=\pi }$
must be taken into account. Alternatively, first evaluate the indefinite integral, then apply the boundary values.
$${\displaystyle {\begin{aligned}\int {\frac {dx}{2+\cos x}}&=\int {\frac {1}{2+{\frac {1-t^{2}}{1+t^{2}}}}}{\frac {2\,dt}{t^{2}+1}}&&t=\tan {\tfrac {x}{2}}\\[6pt]&=\int {\frac {2\,dt}{2(t^{2}+1)+(1-t^{2})}}=\int {\frac {2\,dt}{t^{2}+3}}\\[6pt]&={\frac {2}{3}}\int {\frac {dt}{{\bigl (}t{\big /}{\sqrt {3}}{\bigr )}^{2}+1}}&&u=t{\big /}{\sqrt {3}}\\[6pt]&={\frac {2}{\sqrt {3}}}\int {\frac {du}{u^{2}+1}}&&\tan \theta =u\\[6pt]&={\frac {2}{\sqrt {3}}}\int \cos ^{2}\theta \sec ^{2}\theta \,d\theta ={\frac {2}{\sqrt {3}}}\int d\theta \\[6pt]&={\frac {2}{\sqrt {3}}}\theta +C={\frac {2}{\sqrt {3}}}\arctan \left({\frac {t}{\sqrt {3}}}\right)+C\\[6pt]&={\frac {2}{\sqrt {3}}}\arctan \left({\frac {\tan {\tfrac {x}{2}}}{\sqrt {3}}}\right)+C.\end{aligned}}} $$
By symmetry,
$$\begin{align}
\int_{0}^{2\pi} \frac{dx}{2 + \cos x} &= 2 \int_{0}^{\pi} \frac{dx}{2 + \cos x} = \lim_{b \rightarrow \pi} \frac{4}{\sqrt3} \arctan \left( \frac{\tan\tfrac x2}{\sqrt3}\right) \Biggl|_{0}^{b}\\[6pt]
&= \frac{4}{\sqrt3} \Biggl[ \lim_{b \rightarrow \pi} \arctan \left(\frac{\tan\tfrac b2}{\sqrt3}\right) -  \arctan (0) \Biggl] = \frac{4}{\sqrt 3} \left( \frac{\pi}{2} - 0\right) = \frac{2\pi}{\sqrt 3},
\end{align}$$
From: https://en.wikipedia.org/wiki/Tangent_half-angle_substitution

 [/INST] THASsing, as I like to call the process, is perfectly fine here as there are no poles (which solve $\sin x=2\cos x$ or $\tan x=2$) on the interval of integration. After simplifying you get
$$\int_1^\infty\frac{1+t^2}{2(t^2+t-1)^2}\,dt=\left[-\frac t{2(t^2+t-1)}\right]_1^\infty=\frac12$$
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