What did griffiths do here? Trig Identity for buffons needle I'm on the very last part of the equation for buffons needle.  I think it's a trig identity, but I can't find it. Either way, I can give more info if needed but it looks like this.

 A: The equality holds not because the integrands are necessarily equal, but because $$\int_0^l\sin^{-1}\left(\frac{l-y}{l}\right) dy = \int_0^l\sin^{-1}\left(\frac{y}{l}\right)dy.\qquad (\ast)$$ 
Once you have this equality, then 
\begin{align*}
&\frac{1}{\pi l}\int_0^l\pi - \sin^{-1}\left(\frac{y}{l}\right) - \sin^{-1}\left(\frac{l-y}{l}\right)dy\\ 
= &\frac{1}{\pi l}\left[\int_0^l\pi dy - \int_0^l\sin^{-1}\left(\frac{y}{l}\right)dy - \int_0^l\sin^{-1}\left(\frac{l-y}{l}\right)dy\right]\\
= &\frac{1}{\pi l}\left[\int_0^l\pi dy - \int_0^l\sin^{-1}\left(\frac{y}{l}\right)dy - \int_0^l\sin^{-1}\left(\frac{y}{l}\right)dy\right]\\
= &\frac{1}{\pi l}\int_0^l\pi - \sin^{-1}\left(\frac{y}{l}\right) - \sin^{-1}\left(\frac{y}{l}\right)dy\\
= &\frac{1}{\pi l}\int_0^l\pi - 2\sin^{-1}\left(\frac{y}{l}\right)dy.
\end{align*}
To see why $(\ast)$ is true, let $z = l - y$ in the first integral, then we have $$\int_0^l\sin^{-1}\left(\frac{l-y}{l}\right) dy = \int_l^0\sin^{-1}\left(\frac{z}{l}\right)(-dz) = \int_0^l\sin^{-1}\left(\frac{z}{l}\right)dz.$$ Replacing $z$ by $y$ gives the final result.
