<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] Use the result in this post:

Note that
\begin{align}
\frac{\frac14\sin^2\pi x}{\sin^2\frac{\pi x}2-a^2}
=-a^2 + \cos^2\frac{\pi x}2 + \frac{a^2(1-a^2)}{\sin^2\frac{\pi x}2-a^2}
\end{align}
Integrate respectively to obtain
\begin{align}
&\int_ 0^{x_{\downarrow}} \frac{\sin^2\pi x}{\sin^2\frac{\pi x}2-a^2}dx\\
=& \ 2(1-2a^2) x_{\downarrow} +\frac2\pi\sin\pi x_{\downarrow}
-\frac{8a\sqrt{1-a^2}}\pi\tanh^{-1}\frac{\sqrt{1-a^2} \tan\frac{\pi x_{\downarrow}}2}a
\\\\
&\int^1_{x_{\uparrow}} \frac{\sin^2\pi x}{\sin^2\frac{\pi x}2-a^2}dx\\
=& \ 2(1-2a^2)(1- x_{\uparrow})-\frac2\pi\sin\pi x_{\uparrow}
+\frac{8a\sqrt{1-a^2}}\pi\coth^{-1}\frac{\sqrt{1-a^2} \tan\frac{\pi x_{\uparrow}}2}a
\end{align}

Set $a^2=t$ and take derivative with respect to $t$, and replace $t=a^2$ back. It is done.
Further, by taking the $(n-1)^{\text{th}}$ derivative with respect to $t$, you can generalize it into
$$\int^1_{x_{\uparrow}} \frac{\sin^2(\pi x)}{\left(\sin^2\left(\frac{\pi x}{2}\right)-a^2\right)^n}dx$$
Moreover, by taking the integral with respect to $t$, you can get
$$\int^1_{x_{\uparrow}} \sin^2(\pi x)\cdot \ln\left(\sin^2\left(\frac{\pi x}{2}\right)-a^2\right) dx$$
</s>