Indefinite Integral 2

\hspace{-16}\mathbf{\int\sqrt{\frac{3-x}{3+x}}.\sin^{-1}\left(\sqrt{\frac{3-x}{6}\right)dx}}$\\\\\\ Let $\mathbf{\sqrt{\frac{3-x}{6}}=t\Leftrightarrow \frac{3-x}{6}=t^2=\frac{(3-x)}{(3-x)+(3+x)}=t^2}$\\\\\\ So $\mathbf{\frac{3-x}{3+x}=\frac{t^2}{1-t^2}\Leftrightarrow \sqrt{\frac{3-x}{3+x}}=\frac{t}{\sqrt{1-t^2}}}$\\\\\\ and $\mathbf{dx=-12tdt}$\\\\\\ So Integral Convert into\\\\\\ $\mathbf{=-12\int t^2.\sin^{-1}(t).\frac{1}{\sqrt{1-t^2}}dt}$\\\\\\ $\mathbf{=12\int \sqrt{1-t^2}\sin^{-1}(t)-12\int \sin^{-1}(t).\frac{1}{\sqrt{1-t^2}}dt}$\\\\\\ Now Let $\mathbf{I_{1}=\sqrt{1-t^2}\sin^{-1}(t)dt}$\\\\ Put $\mathbf{t=\sin \theta\Leftrightarrow dt=\cos\theta d\theta}$\\\\ So $\mathbf{I_{1}=\int \theta.\cos^2\theta d\theta}$\\\\\\ $\mathbf{=\frac{\theta^2}{4}+\frac{1}{4}.\theta.\cos (2\theta)+\frac{1}{8}.\cos (2\theta)}$

\hspace{-16}$So $\mathbf{I_{1}=\frac{\sin^{-1} t}{4}+\frac{\sin^{-1} t}{4}.\sin \left(2\sin^{-1}t\right)+\frac{1}{8}.\cos \left(2\sin^{-1}t\right)}$\\\\\\ and $\mathbf{I_{2}=\int \frac{\sin^{-1}t}{\sqrt{1-t^2}}dt=\frac{\sin^{-1}(t)}{2}}$\\\\\\ So $\mathbf{12\int \sqrt{1-t^2}\sin^{-1}(t)-12\int \sin^{-1}(t).\frac{1}{\sqrt{1-t^2}}dt=12I_{1}-12I_{2}}$\\\\\\ $\mathbf{=3\sin^{-1}(t)+3.\sin^{-1}.\sin \left(2\sin^{-1}t\right)+\frac{3}{2}.\cos \left(2\sin^{-1}t\right)-6.\sin^{-1}(t)+C}$\\\\\\ Where $\mathbf{t=\sqrt{\frac{3-x}{6}}}$