another integral

\hspace{-16}$Here $\mathbf{4x^2+4x=(2x+1)^2-1}$\\\\\\ So $\mathbf{\int_{-\frac{1}{2}}^\frac{1}{2}\frac{\sqrt{3-4x-4x^2}}{4x^2+4x+5}dx=\int_{-\frac{1}{2}}^\frac{1}{2}\frac{\sqrt{4-(2x+1)^2}}{4+(2x+1)^2}dx}$\\\\\\ Now Let $\mathbf{2x+1=t\Leftrightarrow dx=\frac{1}{2}dt}$ and Changing Limit....\\\\\\ $\mathbf{=\frac{1}{2}.\int_{0}^{1}\frac{\sqrt{4-t^2}}{4+t^2}}$\\\\\\ Again Put $\mathbf{t=2\sin \theta\Leftrightarrow dt = 2\cos \theta d\theta}$\\\\\\ $\mathbf{=\frac{1}{2}.\int_{0}^{\frac{\pi}{2}}\frac{4.\cos^2\theta}{4.(1+\sin^2 \theta)}d\theta = \frac{1}{2}.\int_{0}^{\frac{\pi}{2}} \frac{\cos ^2\theta}{1+\sin ^2 \theta}d\theta}$\\\\\\ $\mathbf{=\frac{1}{2}.\int_{0}^{\frac{\pi}{2}}\frac{1}{1+\left(\sqrt{2}.\tan \theta\right)^2}d\theta}$\\\\\\ Now $\mathbf{\sqrt{2}.\tan \theta =u\Leftrightarrow \sqrt{2}.\sec^2\theta d\theta = du}$\\\\\\ So $\mathbf{=\frac{1}{2.\sqrt{2}}\int_{0}^{\infty}\frac{2}{(1+u^2)(2+u^2)}du}$\\\\\\ So $\mathbf{=\frac{\pi}{4}.\left(\sqrt{2}-1\right)}$\\\\\\