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?
 A: 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?
A: 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 $\displaystyle tdt = -\frac{1}{3}udu$
$\displaystyle \mathcal{I} = -\int\frac{1}{3-3u^2}du = -\int\frac{1}{u^2-1}du$
A: Take $\cos^6 (x) $ common from denominator then use $\tan (2x)=\frac {2\tan (x)}{1-\tan^2 (x)} $ get  every cos to sec put $tan^2 (x)=t $ thus $2\tan (x).sec^2 (x)dx=dt $ thus integral reduces to $\frac {1}{(1-t)\sqrt{t^2-t+1} }$ after using $t^3+1=(t+1)(t^2-t+1) $ thus pull out $u^2$ common from root after putting $1-t=u $  abd doing $-dt=du $put $\frac {1}{u} =l $ thus $-\frac{1}{u^2}du=dl $ and the integral reduces to a very simple form of integrating $\frac {1}{\sqrt {l^2-l+1}} dl$
