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?

as you write $\displaystyle \sin^6 \theta+\sin^6 \theta = 1-3\sin^2 \theta \cos^2 \theta=\frac{1}{4}\left(4-3\sin^2 2 \theta\right)$
let $\displaystyle \mathcal{I} = \int\frac{\tan 2 \theta }{\sqrt{\sin^6 \theta+\sin^6 \theta}}d\theta = 2\int\frac{\sin 2 \theta }{\cos 2 \theta \sqrt{4-3\sin^2 2 \theta}}d\theta = 2\int\frac{\sin 2 \theta\cos 2 \theta }{\cos^2 2 \theta \sqrt{4-3\sin^2 2 \theta}}d\theta$
substitute $\sin 2 \theta = t$ and $2\cos 2 \theta d \theta = dt$
$\displaystyle \mathcal{I} = \int\frac{t}{(1-t^2)\sqrt{4-3t^2}}dt$
substitute $4-3t^2 = u^2$ and $\displaystyle tdt = -\frac{1}{3}udu$
$\displaystyle \mathcal{I} = -\int\frac{1}{3-3u^2}du = -\int\frac{1}{u^2-1}du$
