<s>[INST] Find $\int_{t}^1 \frac{\sin(\pi x)^2}{(\sin(\pi x/2)^2-a^2)^2} \ dx. $ Let $a \in (0,1)$, I would like to integrate
$$\int_{t}^1 \frac{\sin(\pi x)^2}{(\sin(\pi x/2)^2-a^2)^2} \ dx. $$
Now $\sin(\pi x/2)^2$ is a monotonically increasing function from $0$ to $1$, therefore there exists a unique $t^*$ such that $\sin(\pi t^*/2)^2 = a^2.$ Now $t^* <t$ such that the integral is well-defined.
Please let me know if you have any questions.
 [/INST] With $x\to 1-x$
\begin{align}
&\int_{t}^1 \frac{\sin^2\pi x}{(\sin^2\frac{\pi x}2-a^2)^2} \ dx\\
=&\ 4\int_0^{1-t}-1
 +\frac{(1-2a^2)\cos^2\frac{\pi x}2+a^4}{(\cos^2\frac{\pi x}2-a^2)^2}\ dx\\
=& \ 4(t-1)+\frac2a\frac d{da}\int_0^{1-t} \frac{a^2-a^4}{\cos^2\frac{\pi x}2-a^2}\ dx\\
 =& \ 4(t-1)+\frac2a\frac d{da}\bigg(
 \frac{2 a\sqrt{1-a^2}}{\pi}\tanh^{-1}\frac{a\cot\frac{\pi t}2}{\sqrt{1-a^2}}\bigg)\\
=& \ 4\bigg(t-1
-\frac{\frac1\pi\cot\frac{\pi t}2}{a^2\csc^2 \frac{\pi t}2-1}+ \frac{1-2a^2}{\pi a\sqrt{1-a^2}}
 \tanh^{-1}\frac{a\cot\frac{\pi t}2}{\sqrt{1-a^2}}\bigg)
\end{align}
</s>