Trignometric integration We have to solve the following integration. 
$$\int\frac{\tan 2\theta}{\sqrt{\cos^6 \theta+\sin^6\theta}}\ d\theta$$
In this, I write the denominator as $\sqrt 1-3\sin^2\theta \cos^2\theta$
But now how to proceed?

hint: Put $t = \cos (2\theta)\implies 1 - 3\sin^2\theta\cos^2\theta = 1 - \dfrac{3}{4}\sin^2(2\theta)= \dfrac{1}{4} + \dfrac{3}{4}\cos^2(2\theta)\implies I = -\displaystyle \int\dfrac{dt}{t\sqrt{1+3t^2}}$. From this put again $\sqrt{3}t= \tan(\phi)$. Can you take it from here?
