Помогите продолжить решение

sin⁴t=(sin²t)²=((1-cos2t)/2)²

cos⁴t=(cos²t)²=((1+cos2t)/2)²

sin⁴t+cos⁴t=((1-cos2t)/2)²+((1+cos2t)/2)²=(2+2cos²2t)/4=

16sin²tcos²t=16(sin2t/2)²=4sin²2t

$L=\int\limits^\frac{\pi}{2}_0 {\sqrt{2(1+(cos2t)^2)}}sin2t \, dt =\\ \\= -\frac{\sqrt{2}}{2} \int\limits^\frac{\pi}{2}_0 {\sqrt{1+(cos2t)^2}} \, d(cos2t) = -\frac{\sqrt{2}}{2}ln|cos2t+\sqrt(1+cos^22t)|^\frac{\pi}{2}_0=\\ \\ -\frac{\sqrt{2}}{2}ln|cos\pi+\sqrt(1+cos^2\pi)|+\frac{\sqrt{2}}{2}ln|cos0+\sqrt(1+cos^20)|=\\ =-\frac{\sqrt{2}}{2}ln|-1+\sqrt(2)| +\frac{\sqrt{2}}{2}ln|1+\sqrt(2)|$