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.
 A: Hint
$$I=\int \frac{\sin ^2(\pi  x)}{\left(\sin ^2\left(\frac{\pi  x}{2}\right)-a^2\right)^2} \, dx$$
$$x=\frac{2 }{\pi }u \implies I=\frac 2 \pi \int \frac{ \sin ^2(2 u)}{  \left(\sin ^2(u)-a^2\right)^2}\,du$$
$$u=\cot ^{-1}(v)\implies\int \frac{ \sin ^2(2 u)}{  \left(\sin ^2(u)-a^2\right)^2}\,du=4\int\frac{v^2}{(1+v^2)\left(a^2 \left(v^2+1\right)-1\right)^2}\,dv$$
Now use
$$a^2 \left(v^2+1\right)-1=a^2 \left(v-\frac{\sqrt{1-a^2}}{a}\right) \left(v+\frac{\sqrt{1-a^2}}{a}\right)$$
Then, partial fraction decomposition.
I am sure that you can take from here.
When solved, back from $v$ to $u$ and from $u$ to $x$ and use the bounds.
A: 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
$$\int^1_{x_{\uparrow}} \sin^2(\pi x)\cdot \ln\left(\sin^2\left(\frac{\pi x}{2}\right)-a^2\right) dx$$
A: 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}
