The value of the integral $\int\limits_{0}^{π/4}\frac{\sin x\cos x}{\cos^2x+\sin^4x}dx$ is:

$\frac{π}{\sqrt{3}}$ $\frac{π}{6\sqrt{3}}$ $\frac{π}{4\sqrt{3}}$ None of these

$\frac{π}{6\sqrt{3}}$

Put sin2 x = t; $I=\frac{1}{2}\int\limits_0^{1/2}\frac{dt}{(1-t)+t^2}=\frac{1}{2}\int\limits_0^{1/2}\frac{dt}{(t-\frac{1}{2})^2+(\frac{\sqrt{3}}{2})^2}=\frac{1}{\sqrt{3}}[\tan^{-1}\frac{t-\frac{1}{2}}{\sqrt{3}}×2]_0^{1/2}=\frac{π}{6\sqrt{3}}$