The integral is
$\int_{\sqrt{|T^2-w^2|}}^{T+w} \left(1-I_{\sin^{2}\alpha}\left(\frac{n+1}{2},0.5\right)\right)^{M-1}x^{n-1}I_{\sin^{2}\beta}\left(\frac{n-1}{2},0.5\right)dx$
where $I$ is the imcomplete beta function, $n\in\mathbb{N},M\in\mathbb{N},T\geq0,w\geq0$ are all fixed coefficients, $\alpha,\beta$ are two angles within one triangle whose subtense are $w,T$ (and $x$ is the third side of the triangle), respectively (i.e., $\frac{\sin\alpha}{w}=\frac{\sin\beta}{T}$). So,
$\sin^{2}\alpha=1-\cos^{2}\alpha=1-\left(\frac{x^2+T^2-w^2}{2xT}\right)^2$
$\sin^{2}\beta=1-\cos^{2}\beta=1-\left(\frac{x^2+w^2-T^2}{2xw}\right)^2$
I've been trying this integral for months and attempt to solve it via comparing the expression of $\frac{\partial I_{\sin^{2}\alpha}\left(\frac{n+1}{2},0.5\right)}{\partial x}$ and $I_{\sin^{2}\beta}\left(\frac{n-1}{2},0.5\right)$ as
$\frac{\partial I_{\sin^{2}\alpha}\left(\frac{n+1}{2},0.5\right)}{\partial x}=B^{-1}\left(\frac{n+1}{2},0.5\right)\sin^{n-1}\alpha \frac{\cos\beta}{x}\frac{w^2}{T^2}$
$I_{\sin^{2}\beta}\left(\frac{n-1}{2},0.5\right)=\frac{2}{n-1}B^{-1}\left(\frac{n-1}{2},0.5\right)\sin^{n-1}\beta G\left(\sin\beta\right)=\frac{2}{n-1}B^{-1}\left(\frac{n-1}{2},0.5\right)\sin^{n-1}\alpha \frac{T^{n-1}}{w^{n-1}} G\left(\sin\beta\right)$
where $B$ is the beta function, $G(\sin\beta)=1+\sum_{k=1}^{\infty}\frac{\frac{n-1}{2}}{\frac{n-1}{2}+k}\frac{(0.5)_k}{k!}\sin^{2k}\beta$.
The expression between them are close if some approximation to $G(\sin\beta)$can be made, but still there remains $x^{n-1}$ which makes the integral still not solvable.
Is there anyone who have some idea about this integral? It would also be ok to obtain a close closed-form lower bound (still hard to me).